Important Questions for Chapter 7 - Triangles

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Important questions based on NCERT syllabus for Chapter 7 - Triangles:

Question 1: In Fig. X and Y are two points on equal sides AB and AC of a ΔABC such that AX = AY. Prove that XC = YB.
Tri
Solution:
In Δs AXC and AYB, we have
AX = AY [Given]
∠A = ∠A [Common angle]
AC = AB [Given]
So,
ΔAXC ≅ ΔAYB [SAS congruence criterion]
⇒ XC = YB [Since corresponding parts of congruent triangles are equal]

Question 2: 6. In Fig., AB || DC. If x = 43 y and y = 38 z, find ∠BCD, ∠ABC and ∠BAD?
lines
Solution: Since AB || DC and transversal BD intersects them at B and D respectively.
∠ABD = ∠BDC [Alternate angles]
And ∠CBD = ∠ADB
∠BDC = x∘ and y = 36∘
∠ABD =x∘ and ∠ADB = 36∘ (Given)
But, it is given that,
x = 4/3y and y= 3/8 z
x = 4/3 × 36 and 36 = 3/8z
x = 48 and z = 36 × 8/3 = 96
Now, in BAD, we have
∠BAD + ∠ADB + ∠ABD = 180∘
∠BAD + 36∘ + x∘ = 180∘
∠BAD + 36∘ + 48∘ = 180∘
∠BAD = 96∘
Thus, ∠BCD = z∘ = 96∘,
∠ABC = x∘ + y∘= 48∘ + 36∘ =84∘ and ∠BAD = 96∘

Question 3: 11. The sum of two angles of a triangle is 80∘ and their difference is 20∘. Find all the angles.
Solution: Let ABC be the triangle
Given that,
∠A + ∠B = 80∘ and ∠A - ∠B =20∘
Adding the two equations, we get
∠A = 50∘ and ∠B = 30∘
We know that,
∠A + ∠B + ∠C = 180∘ [Angle sum property]
⇒ ∠C = 100∘