A park in the shape of quadrilateral abcd

Hint: We can draw a diagonal to the non-right-angled sides such that we obtain a right-angled triangle. Then we can find the length of the diagonal using Pythagoras theorem. Then we can find the area of the right-angled triangle by taking the half of the product of the non-hypotenuse sides. Then we can take the other triangle and find its perimeter and then we can find its area using the formula $A = \sqrt $. Then we can add the area of both the triangles to get the area of the park. Complete step by step solution: We can draw a diagram representing the park and label it. We are given that $\angle C = 90^\circ $ , $AB = 9m$ , $BC = 12m$ , $CD = 5m$ and $AD = 8m$ . Now we can draw the diagonal BD such that BCD forms a right-angled triangle. Now we can apply Pythagoras theorem. $ \Rightarrow B $ . As it is valid for any triangle, we can also use it for finding the area of the right-angled triangle.

Complete step by step solution: The angle of C is $90^\circ $. The length of AB is $9\;\;\]. Note: If $\angle C = 90^\circ $, so we will use right angle triangle formula if $\angle C$ is not equal to $90^\circ $ then, right angle triangle formula should not be used instead of which we need to use triangle formula.

Important Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams You are here Important Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams You are here Important Deleted for CBSE Board 2024 Exams Transcript Question 1 A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? Total Area of park = Area ∆ABD + Area ∆BCD Area of ∆BCD Since DC = 5 m and BC = 12 m, ∠C = 90°, Then ΔBCD is a right angled triangle Area of ∆BCD = 1/2 x base x height = 1/2 × 12 × 5 m2 = 30 m2 Area of ∆ABD Area of triangle = √(s(s−a)(s−b)(s −c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Here, a = 8m, b = 9m, c = BD Since, ∠C = 90° So,applying Pythagoras theorem, BD2 = BC2 + CD2 BD2 = (12)2 + (5)2 BD2 = 144 + 25 BD2 = 169 So, BD = √169 Hence, c = BD = 13 m s = (𝑎 + 𝑏 + 𝑐)/2 Area of Δ ABD = √(𝑠(𝑠 −𝑎)(𝑠 −𝑏)(𝑠 −𝑐)) Putting a = 8m, b = 9m, c = 13 m & s = 15 m Area of Δ ABD = √(15(15−8)(15−9)(15−13))m2 = √(15×7×6×2) m2 = √((3×5)×7×6×2) m2 = √((3×2)×6×(7×5)) m2 = √((6)×6×(35)) m2 = √((6)2× (35)) m2 = √62 × √35 = 6√35 m2 = (6 × 5.91) m2 = 35.46 m2 Area of the park = Area of ΔABD + Area of ΔBCD = 35.46 + 30 m2 = ...

Hint: We will find the area of the quadrilateral by dividing the quadrilateral into 2 triangles. We will find the area of each triangle using the formulas and add the areas to get the area of the quadrilateral. Formulas used: We will use the following formulas to solve the question. 1) Pythagoras Theorem: In a right-angled triangle, \[\] The area of the quadrilateral is 65.5 square units. Note: We have used Heron’s formula to find the area of one triangle because the sides are given, so we can easily find the perimeter and then the area. If we have used the formula for the area of the triangle then, we have to create a perpendicular line and then apply Pythagoras theorem to get the height. It would have become a much lengthier process. On the other hand we have used the formula of area of a triangle, because we can easily calculate the height using Pythagoras Theorem. Then using the value we can find the area.

In quadrilateral ABCD, AB= 9m, BC = 12m, CD = 5m and DA = 8m, ∠ C = 90 0 Join BD, Now in right Δ B C D, B D 2 = B C 2 + C D 2 = ( 12 ) 2 + ( 5 ) 2 = 144 + 25 = 169 = ( 13 ) 2 ∴ B D = 13 m Now Area of Δ B C D = 1 2 × 12 × 5 = 30 m 2 and in Δ A B D, s = a + b + c 2 = 9 + 13 + 8 2 = 30 2 = 15 ∴ Area of Δ A B D = √ s ( s − a ) ( s − b ) ( s − c ) = √ 15 ( 15 − 9 ) ( 15 − 13 ) ( 15 − 8 ) = √ 15 × 6 × 2 × 7 = √ 180 × 7 m 2 = √ 36 × 5 × 7 = 6 √ 35 m 2 = 6 ( 5.92 ) = 35.52 m 2 ∴ Total area of quad. ABCD = 30+35.52 m 2 = 65.52 m 2

Given: The figure is given below: The dimensions of the park as follows: Angle C = 90 0, AB = 9m, BC = 12m, CD = 5m And, AD = 8m Now, BD is joined. In by Pythagoras theorem, BD 2 = BC 2 + CD 2 BD 2 = 12 2 + 5 2 BD 2 = 169 BD = 13m Now, Semi perimeter of triangle (ABD) = = 13/2 = 15m Using Heron’s formula, Area of = = = = = m 2 = 35.5 m 2 (approx.) Area of quadrilateral ABCD = Area of = 30 + 35.5 = 65.5 m 2 Thus, the park acquires an area of65.5 m 2

Solution: Let us given that, ∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m From above information we draw the figure, we connected B and D such the the triangle is formed So, the triangle BDC is a right angled triangle By using the Pythagoras theorem we get, BD 2 = BC 2 + CD 2 BD 2 = 12 2 + 5 2 BD 2 = 144 + 25 BD = 169 BD = 13 m Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD Now, Area of ∆BCD = 1/2 × base × height = 1/2 × 12 m × 5 m = 30 m 2 Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m Semi Perimeterof ΔABD s = (a + b + c)/2 = (9 + 8 + 13)/2 = 30/2 = 15 m We know that, Heron's formula Area of triangle = s ( s - a ) ( s - b ) ( s - c ) where, s is the semi-perimeter = half of the perimeter and a, b and c are the sides of the triangle Area of triangle ABD = s ( s - a ) ( s - b ) ( s - c ) = 15 ( 15 - 9 ) ( 15 - 8 ) ( 15 - 13 ) = 15 × 6 × 7 × 2 = 6 35 =35.5 m 2 Area of triangle ABD = 35.5 m 2 Therefore, Area of park ABCD =30 m 2 + 35.5 m 2=65.5 m 2 Hence, the park ABCD occupies an area of 65.5 m 2.