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In a trapezium ABCD, seg AB || seg DC seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 15 | Page 46

solution

According to Pythagoras theorem,
In ∆ABD

\[{AB}^2 = {AD}^2 + {DB}^2 \]
\[ \Rightarrow \left( 25 \right)^2 = \left( 15 \right)^2 + {BD}^2 \]
\[ \Rightarrow 625 = 225 + {BD}^2 \]
\[ \Rightarrow {BD}^2 = 625 - 225\]
\[ \Rightarrow {BD}^2 = 400\]
\[ \Rightarrow BD = 20\]

Now,

\[\text{Area of the triangle ABD} = \sqrt{s\left( s - a \right)\left( s - b \right)\left( s - c \right)}\]
\[s = \frac{a + b + c}{2}\]
\[ = \frac{20 + 25 + 15}{2}\]
\[ = \frac{60}{2}\]
\[ = 30\]
\[\text{Area of the triangle} = \sqrt{30\left( 30 - 25 \right)\left( 30 - 20 \right)\left( 30 - 15 \right)}\]
\[ = \sqrt{30 \times 5 \times 10 \times 15}\]
= 150 sq . units
Also,
\[\text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height}\]
\[ \Rightarrow 150 = \frac{1}{2} \times 25 \times DP\]
\[ \Rightarrow DP = \frac{300}{25}\]
\[ \Rightarrow DP = 12\]

Therefore, height of the trapezium = 12.

Now,
According to Pythagoras theorem,
In ∆ADP

\[{AD}^2 = {AP}^2 + {DP}^2 \]
\[ \Rightarrow \left( 15 \right)^2 = \left( 12 \right)^2 + {AP}^2 \]
\[ \Rightarrow 225 = 144 + {AP}^2 \]
\[ \Rightarrow {AP}^2 = 225 - 144\]
\[ \Rightarrow {AP}^2 = 81\]
\[ \Rightarrow AP = 9\]

∴ AP = QB = 9

∴ CD = PQ = 25 − (9 + 9) = 7

\[\text{Area of Trapezium} = \frac{1}{2} \times \text{Sum of parallel sides} \times \text{Height}\]
\[ = \frac{1}{2} \times \left( 25 + 7 \right) \times 12\]
\[ = \frac{1}{2} \times 32 \times 12\]
\[ = 32 \times 6\]
= 192 sq . units

Hence, A(▢ABCD) = 192 sq. units.

Explanation:-

We know that AB is parallel to DC and that BD is perpendicular to AD. Therefore, angle ABD is a right angle. Similarly, we know that AC is perpendicular to BC. Therefore, angle BAC is a right angle.

Let’s use Pythagoras’ theorem to find the length of BD. We have:

BD² = AD² – AB² BD² = 15² – 25² BD² = 225 – 625 BD² = -400

Since BD cannot be negative, there must be an error in our calculations. This means that our assumption that AB is greater than DC must be incorrect. In fact, DC is greater than AB.

Let’s assume that DC = x. Then we have AB = x + 10 (since AB = 25 and DC = AB – 2BC). We also know that AD = 15 and BC = 15. Using the Pythagorean theorem, we can find BD:

BD² = AD² – AB² BD² = 15² – (x + 10)²

Similarly, we can find AC:

AC² = BC² – CD² AC² = 15² – x²

Since AB is parallel to DC, we have angle ABC = angle ADC. Similarly, since AC is perpendicular to BC, we have angle ABC + angle BAC = 90 degrees. Therefore, we can use trigonometry to find the length of BC in terms of x:

tan(ABC) = BD / BC tan(ADC) = AC / DC tan(ABC) = tan(ADC) BD / BC = AC / DC BD / BC = sqrt(BC² – CD²) / CD BD² = BC² – CD² CD² = BC² – BD² CD = sqrt(BC² – BD²) CD = sqrt(15² – (15² – (x + 10)²)) CD = sqrt((x + 10)² – 15²)

Now we can use the formula for the area of a trapezium:

A(trapezium) = (a + b)h / 2

where a and b are the parallel sides, and h is the height (the perpendicular distance between the parallel sides).

In this case, a = AB = x + 10, b = DC = x, and h = CD.

A(trapezium) = ((x + 10) + x) * sqrt((x + 10)² – 15²) / 2

Simplifying this expression and using the fact that x² + 20x + 75 = (x + 15)(x + 5), we get:

A(trapezium) = (3x + 25) * sqrt(x² + 20x + 75) / 2

To find the area of the trapezium, we need to find the value of x. We can use the fact that the diagonals of a trapezium bisect each other to find x:

BD² + AC² = 2(AD² + BC²) (15² – (x + 10)²) + (15² – x²) = 2(15² + 15²) 225 – x² – 20x – 100 + 225 – x² = 450

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 15 | Page 46