Question

Heya✨✨ solve this question❤⏩

$fst \: plz$
both the questions✨❤

Answer 1

Answer:

36°, 60°, 108°, 156°

Step-by-step explanation:

(1)

Let the angles be 3x, 5x, 9x and 13x.

We know that Sum of angles of quadrilateral is 360°.

⇒ 3x + 5x + 9x + 13x = 360

⇒ 30x =  360

⇒ x = 12°

Then:

⇒ 3x = 36°

⇒ 5x = 60°

⇒ 9x = 108°

⇒ 13x = 156°

Therefore, the angles of quadrilateral are 36°, 60°, 108°, 156°

(2)

Consider the ΔABC and ΔBDA,

⇒ AC = BD{Given}

⇒ AB = BA{Common}

⇒ BC = AD{Opposite sides of parallelogram are equal}

∴ ΔACB ≅ ΔBDA{SAS congruence axiom}

∴ ∠ABC = ∠BAD   ------ (1)

⇒ ∠BAD + ∠ABC = 180° {Sum of angles on the same side of traversal}  -- (2)

From (1) & (2), we have

⇒ ∠BAD = ∠ABC = 90°

∴ ∠A = 90°

Therefore, Parallelogram ABCD is rectangle.

Hope it helps!

Answer 2

1] Given ,

The angles in a quadrilateral are in the ratio of ➡ 3:5:9:13

Let the angles be ➡ 3x:5x:9x:13x

As we know , sum of all the angles of a quadrilateral ➡ 360°

▶3x:5x:9x:13x = 360°

▶ 30x = 360°

▶ x = 360/30

▶ x = 12

Substitute 12 in place of x

➡ First angle = 3x = 3(12) = 36°

➡ Second angle = 5x = 5(12) = 60°

➡ Third angle = 9x = 9(12) = 108°

➡ Fourth angle = 13x = 13(12) = 156°

Therefore the four angles of quadrilateral are ➡ 36° , 60° , 108° , 156° .

2] If the diagonals of a parallelogram are equal , then show it is a rectangle .

Ans ➡ Lets say, ABCD is a parallelogram .

Given ➡ The diagonals AC and BD of parallelogram ABCD are equal in length .

Proof ➡

Consider triangles ABD and ACD.

▶ AC = BD [Given]

▶ AB = DC [opposite sides of a parallelogram]

▶ AD = AD [Common side]

∴ ΔABD ≅ ΔDCA [ SSS ]

∠BAD = ∠CDA [CPCT]

∠BAD + ∠CDA = 180° [Adjacent angles of a parallelogram are supplementary.]

So, ∠BAD and ∠CDA are right angles as they are congruent and supplementary.

Therefore, parallelogram ABCD is a rectangle since a parallelogram with one right interior angle is a rectangle.

Hence Proved .