(i) Not similar
(ii) Similar triangles,
$$ \boxed{x = 2.5} $$
$$ \boxed{330\text{ m}} $$
$$ \boxed{42\text{ m}} $$
Given right triangles QPR and QSR on the same base QR; PR and SQ meet at T.
Consider triangles PTS and RTQ. Angle PTS equals angle RTQ (vertical angles) and angle PST equals angle RQT (angles subtended by the same arc / corresponding angles from the right triangles). Hence ΔPTS ∼ ΔRTQ.
From similarity, PT/ST = TQ/TR. Cross-multiplying gives PT × TR = ST × TQ.
$$ \boxed{ AE=\frac{15}{13} } $$
$$ \boxed{ DE=\frac{36}{13} } $$
$$ \boxed{CA=5.6\text{ cm}} $$
$$ \boxed{AQ=3.25\text{ cm}} $$
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.
$$ \boxed{2.8\text{ cm}} $$
$$ \boxed{2\text{ m}} $$
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:
$$ \boxed{\frac45} $$
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:
$$ \boxed{\frac73} $$
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 | $2\text{ m}$ |
By Basic Proportionality Theorem,
$$ \frac{AD}{DB}=\frac{AE}{EC} $$
So,
$$ \frac{AE}{EC}=\frac34 $$
Let
$$ AE=3x,\ EC=4x $$
Then,
$$ AC=AE+EC=7x $$
$$ 7x=15 $$
$$ x=\frac{15}{7} $$
$$ AE=3x=\frac{45}{7}=6.43\text{ cm} $$
Answer
$$ \boxed{6.43\text{ cm}} $$
Using BPT,
$$ \frac{AD}{DB}=\frac{AE}{EC} $$
$$ \frac{8x-7}{5x-3}=\frac{4x-3}{3x-1} $$
Cross multiplication:
$$ (8x-7)(3x-1)=(5x-3)(4x-3) $$
Solving,
$$ 24x^2-29x+7=20x^2-27x+9 $$
$$ 4x^2-2x-2=0 $$
$$ 2x^2-x-1=0 $$
$$ (2x+1)(x-1)=0 $$
$$ x=1 $$
Answer
$$ \boxed{1} $$
First BC = BQ + QC = 35 + 15 = 50 cm.
Since PQ || DC (and AB || DC), by basic proportionality AP/PD = BQ/QC.
So AP/18 = 35/15 ⇒ AP = 18 × (35/15) = 18 × (7/3) = 42 cm.
Therefore AD = AP + PD = 42 + 18 = 60 cm.
$$ \boxed{DE || BC} $$
$$ \boxed{DE || BC} $$
Let the common side of the rhombus be s. Then BP = BR = s.
Since PQ∥BC, by basic proportionality in ΔABC, AP/AB = AQ/AC. Here AP = AB − s = 12 − s, so AQ = AC·(12 − s)/12.
Since QR∥AB, in ΔACB we have CQ/CA = CR/CB. Here CR = CB − s = 6 − s, so CQ = AC·(6 − s)/6. Hence AQ = AC − CQ = AC·s/6.
Equate the two expressions for AQ: AC·(12 − s)/12 = AC·s/6 ⇒ (12 − s)/12 = s/6 ⇒ 12 − s = 2s ⇒ 3s = 12 ⇒ s = 4.
Therefore PQ = RB = 4 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.
$$ AD^2 = AB \times AF $$
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.
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
$$ \boxed{BD=2.5\text{ cm}} $$
$$ \boxed{DC=3.5\text{ cm}} $$
$$ $$ \boxed{\text{Not a bisector}} $$
$$ $$ \boxed{\text{Bisector}} $$
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.
$$ \boxed{2.1\text{ cm}} $$
Construction outline:
- Draw AB = 5.5 cm.
- Through a point at distance 4 cm from line AB draw a line parallel to AB; the vertex C lies on this parallel at a position making ∠C = 25° with AB.
- At a point on the parallel, construct the angle of 25° to meet the ends of AB at appropriate distances to form ΔABC. Join vertices to complete the triangle.
Given: AB = 5.5 cm, ∠C = 25°, distance from C to AB (altitude) = 4 cm.
- Draw segment AB = 5.5 cm.
- Through any point construct a line l parallel to AB at a distance of 4 cm (do this on the side where you want C). The point C must lie on this line l.
- To locate C so that ∠ACB = 25°: construct the circle through A and B that subtends an angle of 25° at the circumference. To do this, note that the corresponding central angle AOB is 50°. At A, construct an angle of 65° with AB (65° = (180° − 50°)/2) on the same side as line l. At B, construct an angle of 65° with BA on the same side. Let the intersection of these two rays be O.
- With O as centre and OA (or OB) as radius, draw the circle through A and B. This circle consists of all points that subtend 25° at AB.
- The intersection point(s) of this circle with the line l give the possible position(s) of C. Choose the appropriate intersection C and join C to A and B.
Verification: By construction, C lies at distance 4 cm from AB and the arc AB of the constructed circle subtends 25° at C, so ∠ACB = 25°. The required triangle ABC is obtained (two symmetric solutions may exist, one on each side of AB).
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.
Construction outline:
- Draw base PQ of required length 6.8 cm.
- At the midpoint (or the chosen vertex for the vertical angle), construct the vertical angle of 50°.
- From the vertex, draw the angle bisector and mark point D on the base such that PD = 5.2 cm.
- Using these constraints, locate the third vertex so that the bisector meets the base at D. Join vertices to complete ΔPQR.
Note: steps summarise the standard ruler-and-compass approach; precise drawing requires graphical construction on paper.
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 the diagonals are equal, so XZ = YW. Given XZ + YW = 26 ⇒ 2·XZ = 26 ⇒ XZ = 13 cm.
Let XY = l and YZ = b. Then l + b = 17 and diagonal gives l^2 + b^2 = 13^2 = 169.
(l + b)^2 = l^2 + b^2 + 2lb ⇒ 17^2 = 169 + 2lb ⇒ 289 = 169 + 2lb ⇒ 2lb = 120 ⇒ lb = 60.
So l and b are roots of t^2 − 17t + 60 = 0 ⇒ (t − 12)(t − 5) = 0.
Length = 12 cm, Breadth = 5 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 horizontal distance x satisfies x^2 + 4^2 = 5^2 ⇒ x^2 = 25 − 16 = 9 ⇒ x = 3 m.
After moving foot 1.6 m towards the wall, new horizontal distance = 3 − 1.6 = 1.4 m.
New height h satisfies h^2 + 1.4^2 = 5^2 ⇒ h^2 = 25 − 1.96 = 23.04 ⇒ h = 4.8 m.
Increase in height = 4.8 − 4 = 0.8 m.
Let SR = x, so QS = 3x and QR = 4x. Using Pythagoras in the right triangles:
PQ^2 = PS^2 + (3x)^2 = PS^2 + 9x^2
PR^2 = PS^2 + x^2
Then 2·PR^2 = 2PS^2 + 2x^2 and 2·PQ^2 = 2PS^2 + 18x^2.
Subtracting gives 2·PQ^2 − 2·PR^2 = 16x^2 = (4x)^2 = QR^2, so
2·PQ^2 = 2·PR^2 + QR^2.
Let BD = DE = EC = x, so BC = 3x.
Then AE^2 = AB^2 + (2x)^2 = AB^2 + 4x^2.
AC^2 = AB^2 + (3x)^2 = AB^2 + 9x^2.
AD^2 = AB^2 + x^2.
Compute 3·AC^2 + 5·AD^2 = 3(AB^2 + 9x^2) + 5(AB^2 + x^2) = 8AB^2 + 32x^2 = 8(AB^2 + 4x^2) = 8·AE^2.
Thus 8·AE^2 = 3·AC^2 + 5·AD^2.
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:
$$ AB^2=OT^2-r^2 $$
$$ =13^2-5^2 $$
$$ =169-25 $$
$$ =144 $$
$$ AB=12 $$
Therefore, the tangent length is:
$$ \boxed{12\text{ cm}} $$
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 the circle with $OP$ as diameter. 5. Let it meet the given circle at $T_1$ and $T_2$. 6. Join $PT_1$ and $PT_2$.
Length of each tangent:
$$ PT=\sqrt{11^2-4^2}=\sqrt{105}\approx10.25\text{ cm} $$
Required tangents obtained.
$$ \boxed{4\text{ cm}} $$
Construction:
- Draw circle with centre O and radius 3.6 cm. Mark point P such that OP = 7.2 cm (P is external).
- Draw segment OP.
- Draw the circle with OP as diameter (centre at midpoint of OP, radius = OP/2). Any intersection T of this circle with the given circle satisfies ∠OTP = 90°, so PT is tangent at T.
- Join P to each intersection point T to obtain the required tangent(s).
Tangent length: PT = √(OP² − OT²) = √(7.2² − 3.6²) = √(972/25) = 18√3/5 ≈ 6.24 cm.
$$ \boxed{(3)\ \angle B=\angle D} $$
$$ \angle N=180^\circ-(60^\circ+50^\circ) $$
$$ =70^\circ $$
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
$$ \boxed{(4)\ 5\sqrt2\text{ cm}} $$
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
$$ \boxed{(1)\ 1.4\text{ cm}} $$
Angle bisector theorem:
$$ \frac{AB}{AC}=\frac{BD}{DC} $$
$$ \frac8{AC}=\frac63=2 $$
$$ AC=4 $$
Answer
$$ \boxed{(2)\ 4\text{ cm}} $$
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
$$ \boxed{(2)\ 110^\circ} $$
Radius is perpendicular to tangent.
Answer
$$ \boxed{(4)\ 90^\circ} $$
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 |
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