Question

✠Question​​
1. In the figure given above ,ABC is a right angle triangle. Find
• The area of triangle ABC
° The length of AC
▪︎The length of BD correct to two places of decimal

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Answer 1

Required Answer:-

Given in figure:

1. Length of AB = 9cm.
2. Length of BC = 40cm.
3. ΔABC is a right angled triangle, right angled at ∠B
4. ΔABD is a right angled triangle, right angled at ∠D

To find:

1. The Area of triangle ABC.
2. The length of AC.
3. The length of BD correct to two decimal places of division.

Solution:

As ΔABC is a right angled triangle,

★ Area of ΔABC,

= ½ × Product of two legs,

= ½ × 9 × 40

= (9 × 20) cm²

= 180 cm²

★ Hence, area of ΔABC is 180 cm²

★ By Pythagoras Theorem,

➡ (Hypotenuse)² = (Height)² + (Base)²

➡ AC² = AB² + BC²

➡ AC² = 9² + 40²

➡ AC² = 81 + 1600

➡ AC² = 1681

➡ AC = √1681

➡ AC = 41 cm.

★ Hence, the length of AC is 41 cm.

Now, we know that,

★ Area of a triangle = ½ × Base × Height.

Here,

1. Base = AC = 41cm.
2. Height = BD

So, Area of ΔABC is,

= ½ × 41 × BD

As area of the triangle is 180cm², So,

➡ ½ × 41 × BD = 180

➡ BD = 180 × 2 × 1/41

➡ BD = 360/41 cm.

➡ BD = 8.78 cm.

★ Hence, the length of BD correct to two decimal places is 8.78 cm.

Answer:

1. The area of the triangle is 180 cm²
2. Length of AC is 41 cm.
3. The length of BD correct to two decimal places is 8.78 cm.

Answer 2

Answer:

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