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.
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:
Serial number |
Statement |
Reason |
1 |
OD = ED |
D is the midpoint of OE |
2 |
DC = DC |
Common side |
3 |
∠CDE = ∠CDO |
Linear pair and given ∠CDE = 90° |
4 |
△ODC ≡ △EDC |
By RHS criteria |
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:
Serial number |
Statement |
Reason |
1 |
In △MUG and △TUB, MU = TU |
∠3 = ∠4, opposite sides of equal angles |
2 |
UG = UB |
∠1 = ∠2, the side opposite to equal angles are equal |
3 |
∠GUM = ∠BUT |
Vertically opposite angles |
4 |
△MUG ≡ △TUB |
By SAS criteria |
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.