Samacheer Class 7 Maths - Geometry Intext Questions

(iv) 3, 6, 8
(i) Here 5 + 9 = 14 = the measure of the third side.
In a triangle the sum of the measures of any two sides must be greater than the third side.
∴ 5, 9, 14 cannot be the sides of a triangle.
(ii) 7.7.15
Here sum of two sides 7 + 7 = 14 < the measures of the thrid side.
So 1,1, 15 cannot be the sides of a triangles.
(iii) 1,2,4
Here sum of two sides 1 + 2 = 3 < the measure of the third side.
∴ 1, 2, 4 cannot be the sides of a triangle.
(iv) 3, 6, 8
Sum of two sides 3 + 6 = 9 > the third side.
∴ 3, 6, 8 can be the sides of a triangle.

No, we cannot have more than one right angle in a triangle.
Because the sum of three angles of a triangle is 180°.
But if two angles are right angles then their sum itself become 180°.

Sum of three angles of a triangle = 180°
If one angle is right angle (i.e. 90°) .
Sum of other two sides = 180° – 90° = 90°

Yes, it is possible.
If one angle is right angle, then the other two angles will be 45° and 45°.

Exercise 4.2
Try These (Text book Page No. 76)

If ∆ABC ≅ ∆XYZ
\(\overline{A B}\) ≅ \(\overline{X Y}\) – \(\overline{B C}\) ≅ \(\overline{Y Z}\)
\(\overline{A C}\) ≅ \(\overline{X Z}\)
And also
∠A ≅ ∠X – ∠B ≅ ∠Y
∠C ≅ ∠Z

(i) In ∆ABC and ∆XYZ
if \(\overline{A B}\) = \(\overline{X Y}\)
\(\overline{B C}\) = \(\overline{Y Z}\)
\(\overline{A C}\) = \(\overline{X Z}\)
then ∆ABC ≅ ∆XYZ. By the Side – Side -Side Congruency Criterion.

then ∆AB ≅ ∆XYZ.
By Side – Angle – Side Criterion.

then ∆ABC ≅ ∆XYZ.
By Angle – Side – Angle Congruency Critirion.

then by RHS criterion.
∆ABC ≅ ∆XYZ
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Given angle 25°, 65° and 80°.
Sum of the angles = 25° + 65° + 80° = 170° ≠ 180
∴ We cannot draw a triangle with these measures.

In ∆AOB and ∆DOC,
∠AOB = ∠DOC [∵ Vertically opposite angles are equal]
Let ∠AOB = ∠DOC = y
By angle sum property of a triangle we have
∠A + ∠B + ∠AOB = ∠D + ∠C + ∠DOC = 180°
3x + 2x + y = 70° + 30° + y = 180°
5x + y = 100° + y = 180°
Here 5x + y = 100° + y
5x = 100° + y – y
5x = 100°
x = \(\frac{100^{\circ}}{5}\) = 20°
∠A = 3x = 3 × 20 = 60°
∠B = 2x = 2 × 20 = 40°
∠A = 60°
∠B = 40°

Given three angles of the triangles are in the ratio 3 : 5 : 4.
Let the three angle be 3x, 5x and 4x.
By angle sum property of a triangle, we have
3x + 5x + 4x = 180°
12x = 180°
x = \(\frac{180^{\circ}}{12}\)
x = 15°
∴ The angle are 3x = 3 × 15° = 45°
5x = 5 × 15° = 75°
4x = 4 × 15° = 60°
Three angles of the triangle are 45°, 75°, 60°

In ∆RST. Let ∠R = x.
Then given S is ∠10° greater than ∠R
∴ ∠S = x + 10°
Also given ∠T is 5° less then ∠S.
So ∠T = ∠S – 5° = (x + 10)° – 5° = x + 10° – 5°
By angle sum property of triangles, sum of three angles = 180°.

∠R + ∠S + ∠T = 180°
x + x + 10° + x + 5° = 180°
3x + 15° = 180°
3x = 180° – 15°
x = \(\frac{165^{\circ}}{3}\) = 55°
∠R = x = 55°
∠S = x + 10° = 55° + 10° = 65°
∠T = x + 5° = 55° + 5° = 60°
∴ ∠R = 55°
∠S = 65°
∠T = 60°

In ABC, Let ∠A = x,
then ∠B = 3 times ∠A = 3x
∠C = 2 times ∠A = 2x
By angle sum property of a triangles,
Sum of three angles of ∆ABC =180°.
∠A + ∠B + ∠C = 180
x + 3x + 2x = 180°
x (1 + 3 + 2) = 180°
6x = 180°

x = \(\frac{180^{\circ}}{6}\) = 30°
∠A = x = 30°
∠B = 3x = 3 × 30° = 90°
∠C = 2x = 2 × 30° = 60°
∴ ∠A = 30°
∠B = 90°
∠C = 60°

Given in ∆XYZ, ∠X : ∠Z = 5 : 4
Let ∠X = 5x; and ∠Z = 4x given ∠Y = 72°
By the angle sum property of triangles sum of three angles of a triangles is 180°.
∠X + ∠Y + ∠Z = 180°
5x + 72 + 4x = 180°
5x + 4x = 180° – 72°
9x = 108°
x = \(\frac{108^{\circ}}{9}\) = 12°
∠X = 5x = 5 × 12° = 60°
∠Z = 4x = 4 × 12° = 48°
∴ ∠X = 60°
∠Z = 48°

Given in ∆ABC ∠B = 90°
∠A = x + 1
∠B = 2x + 5

By angle sum property of triangles
Sum of three angles of ∆ABC = 180°
∠A + ∠B + ∠C = 180°
(x + 1) + 90° + (2x + 5) = 180°
x + 2x + 1° + 90° + 5° = 180°
3x + 96° = 180°
3x = 180° – 96° = 84°
x = \(\frac{84^{\circ}}{3}\) = 28°
∠A = x + 1 = 28 + 1 = 29
∠C = 2x + 5 = 2 (28) + 5 = 56 + 5 = 61
∴ ∠A = 29°
∠C = 61°

Given ∠N = 90°
MO is extended to P, the exterior angle ∠NOP = 128°
Exterior angle is equal to the sum of interior opposite angles.

∴ ∠M + ∠N = 128°
∠M + 90° = 128°
∠M = 128° – 90°
∠M = 38°
By angle sum property of triangles,
∴ ∠M + ∠N + ∠O = 180°
38° + 90° + ∠O = 180°
∠O = 180° – 128°
∠O = 52°
∴ ∠M = 38° and ∠O = 52°

Exterior angle is equal to the sum of the opposite interior angles.
∠LNO = ∠MLN + ∠LMN

6x – 5 = 100° – x + 2x
6x – 5 + x – 2x = 100°
6x + x – 2x = 100° + 5°
5x = 105°
x = \(\frac{105^{\circ}}{5}\) = 21°
x = 21°

Given ∆ABC ≅ DEF.

(i) Corresponding congruent sides.
\(\overline{A B}\) = \(\overline{D E}\); \(\overline{B C}\) = \(\overline{E F}\); \(\overline{A C}\) = \(\overline{D F}\)
(ii) Corresponding congruent angles.
∠ABC = ∠DEF; ∠BCA = ∠EFD ; ∠CAB = ∠FDE

(i) Let the given triangle be ∆ABC. \(\overline{A D}\) divides ∆ABC into two parts giving ∆ABD and ∆ACD.
In ∆ABD and ∆ACD

\(\overline{A B}\) = \(\overline{A C}\) (given)
\(\overline{B D}\) = \(\overline{A D}\) (common side)
∠BAD = ∠CAD (included angles)
∴ By SAS criterion ∆ABD ≅ ∆ACD.
(ii) Let the given triangles in the figure be ∆ABC and ∆DCB.
In both the triangles

\(\overline{B C}\) = \(\overline{B C}\) (Common side)
\(\overline{A B}\) = \(\overline{D C}\)
\(\overline{A C}\) = \(\overline{B D}\)
∴ By SSS Criterion ∆ABC ≅ ∆DCB
(iii) Let the given triangles be ∆ABC and ∆CDE.

Here \(\overline{A C}\) = \(\overline{C E}\) (given)
∠BAC = ∠DEC (given)
∠ACB = ∠DCE (vertically opposite angles)
Two angles and the included side are equal.
Therefore by ASA criterion ∆ABC ≅ ∆CDE.
(iv) Let the two triangles be ∆XYZ and ∆XYW

Here ∠W = ∠Z = 90°
\(\overline{X Y}\) = \(\overline{X Y}\) (Common Hypothenure)
\(\overline{X W}\) = \(\overline{X Z}\) (given)
By RHS criterion ∆XYZ ≅ ∆XYW
(v) Let the two triangles be ∆ABC and ∆ADC

In both the triangles \(\overline{A C}\) = \(\overline{A C}\) (common sides)
\(\overline{A D}\) = \(\overline{B C}\) (given)
\(\overline{A B}\) = \(\overline{D C}\) (given)
By SSS criterion ∆ABC ≅ ∆ADC.

(i) In the given triangles one angle is equal and a side is common and so equal.

To satisfy ASA criterion one more angle should be equal such that the common side is the included side of both angles of a triangle.
The figure will be as follows.
(ii) In the two given triangles two sides of one triangle is equal to two sides of the other triangle.

To satisfy SSS criterion the third sides mut be equal.
(iii) The given triangles have one side in common. They are right angled tringles.

To satisfy RHS criterion their hypotenuse must be equal.
(iv) In the given triangles two angles of one triangle is equal to two angles of the other triangles?

To satisfy ASA criterion included side of two angles must be equal.
(v) In both the triangles one of their sides are equal.

One of their angles are equal or they are vertically opposite angles.
To satisfy SAS criterion, one more side is to be equal such that the angle is the included of the equal sides.

(i) Given two pair of sides are equal and one side is common to both the triangles.
∴ SSS congruency criterion is used.
(ii) One of the sides and one of the angles are equal.
∴ One more angle is vertically opposite angle and so it is also equal.
ASA criterion is used.
(iii) From the figure hypotenuse and one side are equal in both the triangles.
RHS congruency criterion is used. (∵ Considering ∆ABC and ∆BAD)
∠A = ∠B = 90°
AD = BC
AB = AB (common)
∴ AC = BD (hypotenuse)
(iv) By ASA criterion both triangles are congruent.
(v) By ASA criterion both triangles are congruent. Since two angles in one triangle are equal to two corresponding angles of the other triangle. Again one side is common to both triangle and the side is the included side of the angles.
(vi) Two sides are equal. One angle is vertically opposite angles and one equal.
By SAS criterion both triangles are cogruent.

Construction:
Step 1: Draw a line. Marked Y and Z on the line such that YZ 7.7 cm.
Step 2: With Y as centre drawn an arc of radius 6.4 cm above the line YZ.
Step 3: With Z as centre, drwan an arc or radius 5 cm to intersect arc drawn in steps. Marked the point of intersection as X.
Step 3: Joined YX and ZX. Now XYZ is the required triangle.
(ii) An equilateral triangle of side 7.5 cm


Construction:
Step 1: Drawn a line. Marked X and Y on the line such that XY = 7.5 cm.
Step 2: With X as centre, drawn an arc of radius 7,5 cm above the line XY.
Step 3: With Y as centre, drawn an arc of radius 7.5 cm to intersect arc drawn in steps. Marked the point of intersection as Z.
Step 4: Joined XZ and YZ. Now XYZ in the required triangle.
(iii) An isosceles triangle with equal sides 4.6 cm and third side 6.5 cm


Construction: .
Step 1: Drawn a line. Marked X and Y on the line such that XY = 6.5 cm.
Step 2: With X as centre, drawn an arc of radius 4.6 cm above the line XY
Step 3: with Y as centre, drawn an arc of radius 4.6 cm to intersect arc drawn in steps. Marked the point of intersection as Z.
Step 4: Joined XZ and YZ. Now XYZ is the required triangle.

In an isoscales triangle, angle opposite to equal sides are equal. Let the equal angles be x° and x°.
In a triangle the sum of the three angles is 180°.
x° + x° + 76° = 180°
x° (1 + 1) = 180° – 76° = 104°
2x = 104°
x = \(\frac{104^{\circ}}{2}\) = 52°
x = 52°
∴ Other two angles are 52° and 52°.

Given two angles of the triangle are same and is equal to 46°. If two angles are equal the sides opposite to equal angles are equal. Therefore it will be an isoscales triangle.

Let ∠B is the greater angle then by the given condition ∠B = ∠A + ∠C.
Sum of three angle of a triangle = 180°.

∠A + ∠B + ∠C = 180°.
∠A + (∠A + ∠C) + ∠C) = 180°.
2∠A + 2∠C = 180°
2(∠A + ∠C) = 180°
∠A + ∠C = \(\frac{180^{\circ}}{2}\)
∠B = 90°
∴ One of the angle of the triangle = 90°
It will be a right angled triangle.

Given the exterior angle = 140°
Interior opposite angle are equal.

Let one of the interior opposite angle be x.
Then x + x = 140°.
[∵ Exterior angle = sum of interior opposite angles]
2x = 140°
x = \(\frac{140^{\circ}}{2}\) = 70°
x = 70°
Interior opposite angle = 70°, 70°.
Sum of the three angles of a triangle = 180°.
70° + 70° + Third angle = 180°
140° + Third angle = 180°
Third angle = 180° – 140° = 40°
∴ Interior angle are 40°, 70°, 70°.

When extending the side KL, the exterior angle formed in equal
to the sum of the interior opposite angles.

∠JLX = ∠LJK + ∠LKJ
= 60°+ 40° =100°
Exterior angle formed = 100°

Given ∆MNO ≅ ∆DEF

∴ Corresponding parts of conqruent triangle are congruent.
∠M = ∠D = 60° [given ∠M = 60°]
∠N = ∠E = 45° [given ∠E = 45°]
∠O = ∠F
In triangle MNO, sum of the three angle – 180°.
∠M + ∠N + ∠O = 180°
60° + 45° + ∠O = 180°
105° + ∠O = 180°
∠O = 180° – 105° = 75°
Value of ∠O = 75°

(i) In ∆BAC and ∆DAC
∠BAC = ∠DAC [Given \(\overline{A Z}\) bisects ∠BAD]
∠BCA = ∠DCA[\(\overline{A Z}\) bisects ∠DCB]
AC = AC [∵ common side]
∴ Here AC is the included side of the angles. By ASA criterior, ∆BAC ≅ ∆DAC.
(ii) By (i) ∆BAC ≅ ∆DAC
BA = DA [By CPCTC]
i.e., AB = AD

In ∆FGH and ∆FHI
Given FG = HI
Also, GH = HI [∵ H is the midpoint of GI]

FH = FH [Common]
∴ By S.SS congruency criteria, ∆FGH ≅ ∆FIH. Hence proved.

In ∆ABC and ∆EBD,

AB = EB
BC = BD
∠ABC = ∠EBD [∵ Vertically opposite angles]
By SAS congruency criteria. ∆ABC ≅ ∆EBD.
We know that corresponding parts of congruent triangles are congruent.
∴ ∠BCA ≅ ∠BDE
and ∠BAC ≅ ∠BED
∠BCA ≅ ∠BDE means that alternate interior angles are equal if CD is the transversal to lines AC and DE.
Similarly, if AE is the transversal to AC and DE, we have ∠BAC ≅ ∠BED
Again interior opposite angles are equal.
We can conclude that AC is parallel to DE.

Challenge Problems

Given that BD = BC
∆BDC is on isoscales triangle.
In isoscales triangle, angles opposite to equal sides are equal.
∠BDC = ∠BCD ……(1)
Also ∠BCD + ∠BCX = 180° [∵ Liner Pair]
∠BCD + 115° = 180°
∠BCD = 180° – 115°
∠BCD = 65° [By (1)]
In ∆ADB
∠BAD + ∠ADB = ∠BDC
[∵ BDC is the exterior angle and ∠BAD and ∠ABD are interior opposite angles]
35° + x = 65°
x = 65° – 35°
x = 30°

Construction:
Step 1: Drawn a line. Marked A and B on the line such that AB = 7 cm.
Step 2: At A, drawn a ray AX making an angle of 120° with AB.
Step 3: With A as centre, drawn an arc of radius 6.5 cm to cut the ray AX. Marked the point of intersection as C.
Step 4: Joined BC.
ABC is the required triangle.
(ii) BC = 8 cm, AC = 6 cm and ∠C = 40°.


Construction:
Step 1: Drawn a line. Marked B and C on the line such fhat BC = 8 Cm.
Step 2: At C, drawn a ray CY making an angle of 40° with BC.

Step 3: With C as centre, drawn an arc of radius 6 cm to cut the ray CY, marked the point of intersection as A.
Step 4: Joined AB.
AB is the required triangle.
(iii) An isosceles obtuse triangle with equal sides 5 cm

Construction:
Step 1: Drawn a line. Marked B and C on the line such that BC = 5 cm.
Step 2: At B drawn a ray BY making on obtuse angle 110° with BC.
Step 3: With B as centre, drawn an arc of radius 5 cm to cut ray BY. Marked the point of intersection as C.
Step 4: Joined BC. ABC is the required triangle.

Construction:
Step 1: Drawn a line. Marked P and R on the line such that PR = 7.8 cm.
Step 2: At P, drawn a ray PX making an angle of 60° with PR.
Step 3: At R, drawn another ray RY making an angle of 35° with PR. Mark the point of intersection of the rays PX and RY as Q.
PQR is the required triangle.
(ii) ∠P = 115°, ∠Q = 40° and PQ = 6 cm

Construction:
Step 1: Drawn a line. Marked P and Q on the line such that PQ = 6 cm.
Step 2: At P, drawn O ray PX making an angle of 115° with PQ.
Step 3: At Q, drawn another ray QY making an angle of 40° with PQ. Marked the point of intersection of the rays PX and Q Y as R.
PQR is the required triangle.
(iii) ∠Q = 90°, ∠R = 42° and QR = 5.5 cm

Construction:
Step 1: Drawn a line. Marked Q and R on the line such that QR = 5.5 cm.
Step 2: At Q, drawn a ray QX making an angle of 90° with QR.
Step 3: At R, drawn another ray RY making an angle of 42° QR. Marked the point of intersection of the rays QX and RY as P.
PQR is the required triangle.
Objective Type Questions

(ii) SAS property
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