Samacheer Kalvi 8th Maths Book Solutions Term 1 Chapter 4 | Geometry Answers For Tamil Nadu Board

Samacheer Kalvi 8th Maths Book Solutions Term 1 Chapter 4 – Geometry is available here. The Samacheer Kalvi 8th Maths book answers of Term 1 Chapter 4, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 8th Maths solutions. Here, in this article below please find the Samacheer Kalvi Class 8 Maths Book Term 1 Chapter 4 Questions with Solutions that we have compiled for the students to solve and practice.

Samacheer Kalvi 8th Class Maths Textbook Term 1 Chapter 4 Questions and Solutions

Term 1 Chapter 4 of the Samacheer Kalvi 8th Maths guide will help the students to solve problems related to similar triangles, congruent triangles, construction of quadrilaterals.

Samacheer Kalvi 8th Maths Term 1 Chapter 4: Geometry Book Exercise 4.1 Questions and Solutions

Question 1: From the given figure, prove that △ABC ~ △DEF.

Solution:

From the △ABC,

AB = AC

It is an isosceles triangle.

Angles opposite to equal sides are equal.

∠B = ∠C = 65^o

∠B + ∠C = 65^o + 65^o = 130^o

Sum of 3 angles of a triangle = 180^o

∠A + ∠B + ∠C = 180^o

∠A + 130^o = 180^o

∠A = 180^o – 130^o

∠A = 50^o

From the △DEF, ∠D = 50^o

Sum of the remaining angles = 180^o – 50^o = 130^o

DE = FD

∠D = ∠F

From △ABC and △DEF,

∠E = 130^o / 2 = 65^o

∠A = ∠D = 50^o

∠B = ∠E = 65^o

∠C = ∠F = 65^o

By AAA criterion, △ABC ~ △DEF.

Question 2: In the given figure, if △EAT ~ △BUN, find the measure of all angles.

Solution:

Given that △EAT ~ △BUN,

Corresponding angles are equal,

∠E = ∠B — (1)

∠A = ∠U — (2)

∠T = ∠N — (3)

∠E = x^o

∠A = 2x^o

Sum of three angles of a triangle = 180^o

In △EAT,

x + 2x + ∠T = 180^o

∠T = 180^o – 3x^o — (4)

Also in △BUN,

x + 40^o + x + ∠U = 180^o

2x + 40^o + ∠U = 180^o

∠U = 140^o – 2x^o

Now by (2)

∠A = ∠U

2x = 140 – 2x

4x = 140

x = 35^o

∠A = 2x = 2 * 35 = 70^o = ∠U

∠N = x + 40^o = 35 + 40^o = 75^o

∠E = 35^o = ∠U

Question 3: In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that △ODC ≡ △EDC.

Solution:

Samacheer Kalvi 8th Maths Term 1 Chapter 4: Geometry Book Exercise 4.2 Questions and Solutions

Question 1: The height of a tower is measured by a mirror on the ground at R by which the top of the tower’s reflection is seen. Find the height of the tower.

Solution:

The image and its reflection make similar shapes.

△PQR ~ △STR,

PQ / ST = QR / TR = PR / SR

PQ / ST = QR / TR

h / 8 = 60 / 10

h = [60 / 10] * 8

= 48 feet

Height of the tower = 48 feet

Question 2: In the figure, given that ∠1 = ∠2 and ∠3 = ∠4. Prove that △MUG ≡ △TUB.

Solution:

Question 3: If △WAR ≡ △MOB, name the additional pair of corresponding parts. Name the criterion used by you.

Solution:

Given: △WAR ≡ △MOB,

∠RWA ≡ ∠BMO [Sum of three angles of a triangle is 180^o]

The criteria used here is the angle sum property of triangles.

Samacheer Kalvi 8th Maths Term 1 Chapter 4: Geometry Book Exercise 4.3 Questions and Solutions

Question 1: PLAY, PL= 7 cm, LA = 6 cm, AY= 6 cm, PA = 8 cm and LY = 7 cm.

Solution:

Steps of construction:

1] A line segment PL = 7cm is drawn.

2] Two arcs of radii 8cm and 6cm are drawn by taking P and L as centres. Both the arcs cut at A.

3] Join PA and LA.

4] Two arcs of radii 7cm and 6cm are drawn by taking A and L as centres. Both the arcs cut at Y.

5] Join LY, PY and AY.

6] The required quadrilateral is PLAY.

Question 2: LIKE, LI = 4.2 cm, IK = 7 cm, KE = 5 cm, LK = 6 cm and IE = 8 cm.

Solution:

Steps of construction:

1] A line segment LI = 4.2 cm is drawn.

2] Two arcs of radii 7cm and 6cm are drawn by taking I and L as centres. Both the arcs cut at K.

3] Join LK and IK.

4] Two arcs of radii 8cm and 5cm are drawn by taking I and K as centres. Both the arcs cut at E.

5] Join LE, IE and KE.

6] The required quadrilateral is LIKE.

Question 3: PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120^o.

Solution:

Steps of construction:

1] A line segment PQ = 3.5 cm is drawn.

2] ∠Q = 120^o is made. Ray QX is drawn.

3] An arc of radius 3.5cm is drawn with Q as the centre and it cuts the ray QX at R.

4] Two arcs of radii 5.2cm and 5.5cm are drawn with R and P as centres which cuts at S.

5] Join PS and RS.

6] The required quadrilateral is PQRS.