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# Solution exercise 9.1| Class seven| Geometry

## Problem-1: In the figure, ∆ABC is a triangle in which ∠ABC = 900, ∠BAC =480 and BD is perpendicular to AC. Find the remaining angles.

Solution: Let remaining angles ∠ABD = x, ∠DBC = y and ∠BCD = z.

Since BD⊥AC

∴ ∠ADB = ∠CDB = 900

Now, in ∆ABD

⟹x+900+480=1800

⟹x+1380=1800

⟹x=1800 – 1380

⟹x=420.

Again, ∠ABC = 900

∴ y +x = ∠ABC = 900

⟹ y+420=900

⟹ y = 900 – 420

⟹ y = 480

And in ∆BCD,

∠CDB+ y + z = 1800

⟹ 900 + 480 + z = 1800

⟹ z + 1380 = 1800

⟹ z = 1800 – 1380

⟹ z = 420

∴ x=420, y = 480, z = 420.

### Problem-2: The vertical angle of an isosceles triangle is 500. Find the other two angles.

Solution: Let, ∆ABC be isosceles triangle where AB = AC and vertical angle ∠BAC = 500. We have to find out the values of other two angles ∠ABC and ∠ACB.

Since in the ∆ABC, AB = AC

∴∠ACB = ∠ABC     [opposite angles are of equal arms are equal]

Let, ∠ACB = ∠ABC = x.

Now, we know that the sum of the three angles of a triangle is 1800.

So, ∠ACB + ∠ABC+ ∠BAC = 1800

⟹ x + x + 500 = 1800

⟹ 2x + 500 = 1800

⟹ 2x = 1800 – 500

⟹ 2x = 1300

⟹ x = 650

∴ ∠ACB = 650 and ∠ABC = 650

# Problem-3: Prove that the sum of the angles of a quadrilateral is equal to 4 right angles.

Solution: General enunciation: We have to prove that the sum of the angles of a quadrilateral is equal to 4 right angles.

Particular enunciation: Let four angles of a quadrilateral ABCD be ∠A, ∠B, ∠C, and ∠D. We have to prove that sum of the four angles equal to 4 right angles that is ∠A+∠B+∠C+∠D= 4 right angles.

Construction: We join A with C and B with D.

Proof: We know the sum of the three angles of a triangle is equal to two right angles.

From triangle ABC

∠BAC+∠ABC+∠BCA=1800—————————————— (1)

Now adding (1) and (2) we have,