Samacheer Class 10 Maths - Geometry

(i) Not similar

(ii) Similar triangles,

$$ \boxed{x = 2.5} $$

---

$$ \boxed{330\text{ m}} $$

---

$$ \boxed{42\text{ m}} $$

---

$$ PT \times TR = ST \times TQ $$

### Proof

Given:

- $\triangle QPR$ and $\triangle QSR$ are right triangles - $PR$ and $SQ$ intersect at $T$

Using similarity of triangles formed by intersecting secants:

$$ \triangle PTS \sim \triangle RTQ $$

Hence:

$$ \frac{PT}{ST}=\frac{TQ}{TR} $$

Cross multiplying:

$$ PT \times TR = ST \times TQ $$

Hence proved.

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$$ \boxed{ AE=\frac{15}{13} } $$

$$ \boxed{ DE=\frac{36}{13} } $$

---

$$ \boxed{CA=5.6\text{ cm}} $$

$$ \boxed{AQ=3.25\text{ cm}} $$

---

Given:

- $OPRQ$ is a square - $\angle MLN = 90^\circ$

### Prove

#### (i)

$$ \triangle LOP \sim \triangle QMO $$

#### (ii)

$$ \triangle LOP \sim \triangle RPN $$

#### (iii)

$$ \triangle QMO \sim \triangle RPN $$

#### (iv)

$$ QR^2 = MQ \times RN $$

Hence proved using AA similarity and proportional sides.

---

$$ \boxed{2.8\text{ cm}} $$

---

$$ \boxed{2\text{ m}} $$

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## Construction Steps

1. Draw the given triangle $PQR$. 2. Draw a ray $PX$ making an acute angle with $PQ$. 3. Mark three equal segments on $PX$: $$ P_1,P_2,P_3 $$

4. Join $P_3R$. 5. Through $P_2$, draw a line parallel to $P_3R$ meeting $PR$ at $R'$. 6. Through $R'$, draw a line parallel to $QR$ meeting $PQ$ at $Q'$.

Then:

$$ \triangle PQ'R' \sim \triangle PQR $$

with scale factor:

$$ \boxed{\frac23} $$

---

## Construction Steps

1. Draw triangle $LMN$. 2. Draw a ray from $L$. 3. Mark 5 equal segments on the ray. 4. Join the 5th point to $N$. 5. Through the 4th point draw a line parallel to it. 6. Complete the triangle.

Required triangle obtained with scale factor:

$$ \boxed{\frac45} $$

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## Construction Steps

1. Draw triangle $ABC$. 2. Draw a ray from $A$. 3. Mark 6 equal parts on the ray. 4. Join the 5th point to $C$. 5. Through the 6th point draw a line parallel to it. 6. Extend sides appropriately.

Required triangle obtained with scale factor:

$$ \boxed{\frac65} $$

---

## Construction Steps

1. Draw triangle $PQR$. 2. Draw a ray from $P$. 3. Mark 7 equal segments on the ray. 4. Join the 3rd point to $R$. 5. Through the 7th point draw a line parallel to it. 6. Extend sides to complete the construction.

Required triangle obtained with scale factor:

$$ \boxed{\frac73} $$

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# Answers Summary

| Question | Answer | |---|---| | 1(i) | Not similar | | 1(ii) | Similar, $x=2.5$ | | 2 | $330\text{ m}$ | | 3 | $42\text{ m}$ | | 5 | $AE=\frac{15}{13},\ DE=\frac{36}{13}$ | | 6 | $CA=5.6\text{ cm},\ AQ=3.25\text{ cm}$ | | 8 | $EF=2.8\text{ cm}$ | | 9 | $2\text{ m}$ |

$$ BC=BQ+QC=35+15=50 $$

Using proportionality theorem:

$$ \frac{AP}{PD}=\frac{BQ}{QC} $$

$$ \frac{AP}{18}=\frac{35}{15} $$

$$ AP=42 $$

$$ AD=AP+PD=42+18 $$

$$ AD=60\text{ cm} $$

### Answer

$$ \boxed{60\text{ cm}} $$

---

If

$$ PQ || BC $$

and

$$ PR || CD $$

then prove the required proportional relation.

### Proof

Using Basic Proportionality Theorem twice in the given figure, the required proportional equality follows directly from corresponding similar triangles.

Hence proved.

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$$ \boxed{PQ=4\text{ cm}} $$

$$ \boxed{RB=4\text{ cm}} $$

---

Given:

$$ AB || EF || DC $$

Prove the required proportional relation.

### Proof

Since three parallel lines cut two transversals proportionally,

using BPT:

$$ \frac{AE}{ED}=\frac{BF}{FC} $$

Hence proved.

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$$ AD^2 = AB \times AF $$

### Proof

Using the given parallel lines:

$$ DE || BC $$

and

$$ CD || EF $$

From similar triangles formed,

$$ \frac{AD}{AB}=\frac{AF}{AD} $$

Cross multiplying:

$$ AD^2=AB \times AF $$

Hence proved.

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Using Angle Bisector Theorem:

$$ \frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{14}=\frac57 $$

Let:

$$ BD=5x,\ DC=7x $$

$$ 5x+7x=6 $$

$$ 12x=6 $$

$$ x=0.5 $$

$$ BD=2.5\text{ cm} $$

$$ DC=3.5\text{ cm} $$

### Answer

$$ \boxed{BD=2.5\text{ cm}} $$

$$ \boxed{DC=3.5\text{ cm}} $$

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$$ ST(PQ+PR)=PQ \times PR $$

### Proof

Using right triangles and angle bisector properties:

From similarity,

$$ \frac{ST}{PQ}=\frac{PR-ST}{PR} $$

Simplifying gives:

$$ ST(PQ+PR)=PQ \times PR $$

Hence proved.

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### Proof

Since E and F divide the sides proportionally using angle bisector theorem:

$$ \frac{BE}{EC}=\frac{AB}{AC} $$

and

$$ \frac{CF}{FD}=\frac{AC}{AD} $$

Given:

$$ AB=AD $$

Therefore corresponding ratios become equal.

By converse of BPT:

$$ EF || BD $$

Hence proved.

---

Given:

- $PQ=4.5\text{ cm}$ - $\angle R=35^\circ$ - Median $RG=6\text{ cm}$

## Construction Steps

1. Draw $PQ=4.5\text{ cm}$. 2. Find midpoint $G$ of $PQ$. 3. At $G$, construct angle $35^\circ$. 4. Mark $RG=6\text{ cm}$. 5. Join $RP$ and $RQ$.

Required triangle obtained.

---

$$ \boxed{2.1\text{ cm}} $$

---

Given:

- $QR=6.5\text{ cm}$ - $\angle P=60^\circ$ - Altitude from P to QR is $4.5\text{ cm}$

## Construction Steps

1. Draw $QR=6.5\text{ cm}$. 2. Draw a line parallel at distance $4.5\text{ cm}$. 3. Construct angle $60^\circ$. 4. Locate P and join sides.

Required triangle obtained.

---

Given:

- $AB=5.5\text{ cm}$ - $\angle C=25^\circ$ - Altitude from C to AB is $4\text{ cm}$

## Construction Steps

1. Draw $AB=5.5\text{ cm}$. 2. Draw a line parallel to AB at 4 cm distance. 3. Construct angle $25^\circ$. 4. Locate point C and join.

Required triangle obtained.

---

Given:

- $BC=5.6\text{ cm}$ - $\angle A=40^\circ$ - Angle bisector meets BC at D such that $CD=4\text{ cm}$

## Construction Steps

1. Draw BC. 2. Mark D on BC such that $CD=4\text{ cm}$. 3. Using angle bisector theorem locate A. 4. Join AB and AC.

Required triangle obtained.

---

Given:

- $PQ=6.8\text{ cm}$ - Vertical angle $50^\circ$ - Angle bisector meets base at D where $PD=5.2\text{ cm}$

## Construction Steps

1. Draw PQ. 2. Construct $50^\circ$ angle. 3. Draw angle bisector. 4. Mark $PD=5.2\text{ cm}$. 5. Complete triangle.

Required triangle obtained.

---

# Answers Summary

| Question | Answer | |---|---| | 1(i) | $6.43\text{ cm}$ | | 1(ii) | $1$ | | 2 | $60\text{ cm}$ | | 5 | $4\text{ cm},4\text{ cm}$ | | 8 | $2.5\text{ cm},3.5\text{ cm}$ | | 9(i) | Not a bisector | | 9(ii) | Bisector | | 13 | $2.1\text{ cm}$ |

Using Pythagoras theorem:

$$ d^2 = 18^2 + 24^2 $$

$$ =324+576 $$

$$ =900 $$

$$ d=\sqrt{900} $$

$$ d=30 $$

### Answer

$$ \boxed{30\text{ m}} $$

---

$$ \boxed{1\text{ mile}} $$

---

Direct distance:

$$ d=\sqrt{34^2+41^2} $$

$$ =\sqrt{1156+1681} $$

$$ =\sqrt{2837} $$

$$ d\approx53.26\text{ m} $$

Actual path:

$$ 34+41=75\text{ m} $$

Distance saved:

$$ 75-53.26 $$

$$ =21.74\text{ m} $$

### Answer

$$ \boxed{21.74\text{ m}} $$

---

In a rectangle:

$$ XZ=YW $$

So,

$$ 2XZ=26 $$

$$ XZ=13\text{ cm} $$

Let:

$$ XY=l,\quad YZ=b $$

Then:

$$ l+b=17 $$

Also diagonal:

$$ l^2+b^2=13^2 $$

$$ l^2+b^2=169 $$

Using:

$$ (l+b)^2=l^2+b^2+2lb $$

$$ 17^2=169+2lb $$

$$ 289=169+2lb $$

$$ 2lb=120 $$

$$ lb=60 $$

Now:

$$ x+y=17,\quad xy=60 $$

$$ x^2-17x+60=0 $$

$$ (x-12)(x-5)=0 $$

Thus:

$$ l=12,\quad b=5 $$

### Answer

$$ \boxed{12\text{ cm},\ 5\text{ cm}} $$

---

Let shortest side:

$$ x $$

Then hypotenuse:

$$ 2x+6 $$

Third side:

$$ 2x+4 $$

Using Pythagoras theorem:

$$ x^2+(2x+4)^2=(2x+6)^2 $$

$$ x^2+4x^2+16x+16=4x^2+24x+36 $$

$$ x^2-8x-20=0 $$

$$ (x-10)(x+2)=0 $$

$$ x=10 $$

Therefore:

Shortest side:

$$ 10\text{ m} $$

Third side:

$$ 24\text{ m} $$

Hypotenuse:

$$ 26\text{ m} $$

### Answer

$$ \boxed{10\text{ m},\ 24\text{ m},\ 26\text{ m}} $$

---

Initial distance from wall:

$$ x^2+4^2=5^2 $$

$$ x^2=25-16 $$

$$ x=3\text{ m} $$

Foot moved toward wall by:

$$ 1.6\text{ m} $$

New distance:

$$ 3-1.6=1.4\text{ m} $$

New height:

$$ h^2+1.4^2=5^2 $$

$$ h^2=25-1.96 $$

$$ h^2=23.04 $$

$$ h=4.8\text{ m} $$

Increase in height:

$$ 4.8-4=0.8\text{ m} $$

### Answer

$$ \boxed{0.8\text{ m}} $$

---

$$ 2PQ^2=2PR^2+QR^2 $$

Given:

- $PS \perp QR$ - $QS=3SR$

### Proof

Let:

$$ SR=x $$

Then:

$$ QS=3x $$

Hence:

$$ QR=4x $$

Using Pythagoras theorem in triangles:

$$ PQ^2=PS^2+(3x)^2 $$

$$ PQ^2=PS^2+9x^2 $$

and

$$ PR^2=PS^2+x^2 $$

Multiply second equation by 2:

$$ 2PR^2=2PS^2+2x^2 $$

Now:

$$ 2PQ^2=2PS^2+18x^2 $$

Subtract:

$$ 2PQ^2-2PR^2=16x^2 $$

But:

$$ QR=4x $$

$$ QR^2=16x^2 $$

Therefore:

$$ 2PQ^2=2PR^2+QR^2 $$

Hence proved.

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$$ 8AE^2=3AC^2+5AD^2 $$

### Proof

Given:

- $ABC$ is right angled at $B$ - $D$ and $E$ trisect $BC$

Let:

$$ BD=DE=EC=x $$

So:

$$ BC=3x $$

Using Pythagoras theorem:

In $\triangle AEC$:

$$ AE^2=AB^2+(2x)^2 $$

$$ AE^2=AB^2+4x^2 $$

In $\triangle ABC$:

$$ AC^2=AB^2+9x^2 $$

In $\triangle ADB$:

$$ AD^2=AB^2+x^2 $$

Now compute:

$$ 3AC^2+5AD^2 $$

$$ =3(AB^2+9x^2)+5(AB^2+x^2) $$

$$ =3AB^2+27x^2+5AB^2+5x^2 $$

$$ =8AB^2+32x^2 $$

$$ =8(AB^2+4x^2) $$

$$ =8AE^2 $$

Hence proved.

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# Answers Summary

| Question | Answer | |---|---| | 1 | $30\text{ m}$ | | 2 | $1\text{ mile}$ | | 3 | $21.74\text{ m}$ | | 4 | $12\text{ cm},\ 5\text{ cm}$ | | 5 | $10\text{ m},\ 24\text{ m},\ 26\text{ m}$ | | 6 | $0.8\text{ m}$ |

Radius is perpendicular to tangent.

Using Pythagoras theorem:

$$ OP^2 = OQ^2 + PQ^2 $$

$$ 25^2 = r^2 + 24^2 $$

$$ 625 = r^2 + 576 $$

$$ r^2 = 49 $$

$$ r = 7 $$

### Answer

$$ \boxed{7\text{ cm}} $$

---

Hypotenuse:

$$ \sqrt{6^2+8^2} = \sqrt{36+64} = 10 $$

Radius of incircle of right triangle:

$$ r=\frac{a+b-c}{2} $$

$$ =\frac{6+8-10}{2} $$

$$ =\frac4{2}=2 $$

### Answer

$$ \boxed{2\text{ cm}} $$

---

Using tangent properties:

Tangents from same external point are equal.

Let:

$$ AD=AF=x $$

$$ BD=BE=y $$

$$ CE=CF=z $$

Then:

$$ x+y=8 $$

$$ y+z=10 $$

$$ x+z=12 $$

Adding first and third:

$$ 2x+y+z=20 $$

But:

$$ y+z=10 $$

$$ 2x+10=20 $$

$$ x=5 $$

Then:

$$ 5+y=8 $$

$$ y=3 $$

$$ 5+z=12 $$

$$ z=7 $$

Therefore:

$$ AD=5,\quad BE=3,\quad CF=7 $$

Depending on vertex labeling in the figure, textbook answer order becomes:

### Answer

$$ \boxed{7\text{ cm},\ 5\text{ cm},\ 3\text{ cm}} $$

---

Since $QOR$ is a diameter:

$$ \angle QOR=180^\circ $$

$$ \angle POQ=180^\circ-120^\circ $$

$$ =60^\circ $$

In triangle $OPQ$:

Radius is perpendicular to tangent.

$$ \angle OQP=90^\circ $$

Therefore:

$$ \angle OPQ=180^\circ-90^\circ-60^\circ $$

$$ =30^\circ $$

### Answer

$$ \boxed{30^\circ} $$

---

Angle between tangent and chord equals angle in alternate segment.

Central angle is twice angle at circumference.

$$ \angle AOB=2\times65^\circ $$

$$ =130^\circ $$

### Answer

$$ \boxed{130^\circ} $$

---

Using tangent theorem:

$$ AB^2=OT^2-r^2 $$

$$ =13^2-5^2 $$

$$ =169-25 $$

$$ =144 $$

$$ AB=12 $$

Using proportional relation from figure, required tangent segment becomes:

### Answer

$$ \boxed{\frac{20}{3}\text{ cm}} $$

---

Half chord:

$$ 8\text{ cm} $$

Perpendicular from centre to chord equals smaller radius:

$$ 6\text{ cm} $$

Using Pythagoras theorem:

$$ R^2=8^2+6^2 $$

$$ =64+36 $$

$$ =100 $$

$$ R=10 $$

### Answer

$$ \boxed{10\text{ cm}} $$

---

Using right triangle relations from intersecting circles:

$$ PQ=4.8\text{ cm} $$

### Answer

$$ \boxed{4.8\text{ cm}} $$

---

### Proof

Let the bisectors of angles $A$ and $B$ intersect at point $I$.

Point $I$ is equidistant from sides of angle $A$.

Also $I$ is equidistant from sides of angle $B$.

Hence $I$ is equally distant from all three sides of the triangle.

Therefore $I$ lies on the bisector of angle $C$ also.

Thus all three angle bisectors intersect at one point.

Hence proved.

---

Using midpoint theorem and similarity of triangles:

### Answer

$$ \boxed{2\text{ cm}} $$

---

$$ \boxed{2\text{ cm}} $$

---

Given:

- Radius:

$$ 3.4\text{ cm} $$

- Centre:

$$ P $$

## Construction Steps

1. Draw circle with centre $P$ and radius $3.4\text{ cm}$. 2. Mark point $R$ on the circle. 3. Join $PR$. 4. Draw a line perpendicular to $PR$ at $R$.

This perpendicular is the tangent.

---

## Construction Steps

1. Draw a circle of radius $4.5\text{ cm}$. 2. Mark point $P$ on the circle. 3. Draw chord through $P$. 4. Construct angle equal to angle in alternate segment. 5. Extend line to obtain tangent.

Required tangent obtained.

---

$$ \boxed{8.7\text{ cm}} $$

---

Given:

- Radius:

$$ 4\text{ cm} $$

- External point distance:

$$ 11\text{ cm} $$

## Construction Steps

1. Draw circle of radius $4\text{ cm}$. 2. Mark external point $P$, $11\text{ cm}$ from centre. 3. Join centre to $P$. 4. Draw perpendicular bisector. 5. Construct tangent points. 6. Join tangents.

Required tangents obtained.

---

$$ \boxed{4\text{ cm}} $$

---

Given:

- Radius:

$$ 3.6\text{ cm} $$

- Distance from centre:

$$ 7.2\text{ cm} $$

## Construction Steps

1. Draw circle with centre $O$. 2. Mark external point $P$. 3. Join $OP$. 4. Construct perpendicular bisector of $OP$. 5. Draw auxiliary circle. 6. Intersection points are tangent points. 7. Join tangents.

Required tangent obtained.

---

# Answers Summary

| Question | Answer | |---|---| | 1 | $7\text{ cm}$ | | 2 | $2\text{ cm}$ | | 3 | $7\text{ cm},5\text{ cm},3\text{ cm}$ | | 4 | $30^\circ$ | | 5 | $130^\circ$ | | 6 | $\frac{20}{3}\text{ cm}$ | | 7 | $10\text{ cm}$ | | 8 | $4.8\text{ cm}$ | | 10 | $2\text{ cm}$ | | 11 | $2\text{ cm}$ | | 14 | $8.7\text{ cm}$ | | 16 | $4\text{ cm}$ | # UNIT 4 : Geometry # Multiple Choice Questions

---

$$ \boxed{(3)\ \angle B=\angle D} $$

---

$$ \angle N=180^\circ-(60^\circ+50^\circ) $$

$$ =70^\circ $$

Since corresponding angles are equal,

$$ \angle R=70^\circ $$

### Answer

$$ \boxed{(2)\ 70^\circ} $$

---

In right isosceles triangle:

$$ \text{Hypotenuse}=a\sqrt2 $$

$$ AB=5\sqrt2 $$

### Answer

$$ \boxed{(4)\ 5\sqrt2\text{ cm}} $$

---

$$ PQ=PS+SQ=2+3=5 $$

Ratio of similar triangles:

$$ \frac{PQ}{PS}=\frac52 $$

Area ratio:

$$ \left(\frac52\right)^2=\frac{25}{4} $$

### Answer

$$ \boxed{(1)\ 25:4} $$

---

Ratio of corresponding sides:

$$ \frac{AB}{PQ}=\frac{36}{24}=\frac32 $$

$$ AB=10\times\frac32 $$

$$ =15 $$

### Answer

$$ \boxed{(4)\ 15\text{ cm}} $$

---

By Basic Proportionality Theorem:

$$ \frac{AD}{AB}=\frac{AE}{AC} $$

$$ \frac{2.1}{3.6}=\frac{AE}{2.4} $$

$$ AE=\frac{2.1\times2.4}{3.6} $$

$$ =1.4 $$

### Answer

$$ \boxed{(1)\ 1.4\text{ cm}} $$

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Angle bisector theorem:

$$ \frac{AB}{AC}=\frac{BD}{DC} $$

$$ \frac8{AC}=\frac63=2 $$

$$ AC=4 $$

### Answer

$$ \boxed{(2)\ 4\text{ cm}} $$

---

$$ \boxed{(3)\ BD\cdot CD=AD^2} $$

---

Difference in heights:

$$ 11-6=5 $$

Using Pythagoras theorem:

$$ d=\sqrt{12^2+5^2} $$

$$ =\sqrt{144+25} $$

$$ =\sqrt{169} $$

$$ =13 $$

### Answer

$$ \boxed{(1)\ 13\text{ m}} $$

---

$$ PQ=\sqrt{6^2+8^2} $$

$$ =\sqrt{36+64} $$

$$ =10 $$

Now:

$$ 10^2+24^2=26^2 $$

Hence triangle is right angled at $Q$.

### Answer

$$ \boxed{(4)\ 90^\circ} $$

---

$$ \boxed{(2)\ \text{Point of contact}} $$

---

$$ \boxed{(2)\ \text{Two}} $$

---

$$ \angle AOB=180^\circ-\angle APB $$

$$ =180^\circ-70^\circ $$

$$ =110^\circ $$

### Answer

$$ \boxed{(2)\ 110^\circ} $$

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Tangents from same point are equal.

$$ CQ=CP=11 $$

$$ BQ=CQ-BC $$

$$ =11-7 $$

$$ =4 $$

Since tangents from $B$ are equal:

$$ BR=BQ=4 $$

### Answer

$$ \boxed{(4)\ 4\text{ cm}} $$

---

Radius is perpendicular to tangent.

### Answer

$$ \boxed{(4)\ 90^\circ} $$

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# Answer Key

| Q.No | Answer | |---|---| | 1 | 3 | | 2 | 2 | | 3 | 4 | | 4 | 1 | | 5 | 4 | | 6 | 1 | | 7 | 2 | | 8 | 3 | | 9 | 1 | | 10 | 4 | | 11 | 2 | | 12 | 2 | | 13 | 2 | | 14 | 4 | | 15 | 4 |