# 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.

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?

*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∘