PYQs for Class 9 Math Section C - Salwan Public School Gurgaon

Solution

Given BE = CD

When 2 sides of a triangle are equal then it is isosceles

When 2 angles and side of 2 angles are equal, then both triangles are simillar

∠ABP = ∠AQB (each 90°)

In ∆ ABE and ∆ ACD

BE = CD (Given)

∠BEA = ∠CDA (each 90°)

∠BAE = ∠CAD (Common angle)

=> ABE = ∠ACE (By angle sum property)

=> ∆ ABE is simillar to ∆ ACD

=> AB = AC

∴ The triangle is isosceles

Solution

Given: L is the bisector of an angle A and BT ⊥ AP and PQ

Proof: ∠BAP = ∠BAQ (L is the bisector of ∠A) -> given

∠ABP = ∠AQB (each 90°)

(given BP and BQ are perpendiculars from B to the area of angle A)

AB = BA (Common)

∴ ∆ APB ≅ ∆ AQB (∵ by ∆ AS)

=> BP = BQ (by CPCT)

or Point B is equidistant from area of ∆ A

Solution

√6 + √3 √6 - √3

√6 + √3 √6 - √3    x      √6 + √3 √6 + √3

(√6 + √3)² 6 - 3

6 + 3 + 2√6 x √3 3

9 + 2√6x3 3

9 + 2√3x2x3 3

9 + 2 x 3√2 3

= 3 + 2√2

= a + b√2

=> a = 3 and b = 2

Solution

Given: Two parallel lines let l ∥ m

Let xy is a transversal intersecting l at p and m at Q.

PM and QN are bisectors of ∠XPL and QN is bisector of ∠PQM and ∠XPL = ∠PQM are corresponding ∠'s.

To prove: PM ∥ QN

Proof: ∠XPL = ∠PQM (Corresponding ∠'s as l ∥ m -> given)

=> ∠XPL / 2 = ∠PQM / 2

=> ∠1 = ∠3 (∵ PM is bisector of ∠XPL ∴ ∠1 = ∠2 and QN is bisector of ∠PQM ∴ ∠1 = ∠3)

∵ corresponding ∠'s are equal ∴ PM ∥ QN

Hence proved