Measuring Space: Perimeter and Area | Exercise 6.2

Question 1

Find the area of triangle ADE in Fig. 6.31.

Question diagram 1
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Solution
Understand the Question
  • To find the area of triangle ADEADE, we use the formula: Area=12×base×height\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}
  • From the figure, the base is side ADAD, and the perpendicular height corresponds to the distance from vertex EE to the opposite side of the rectangle containing base ADAD.

Step 1 · Identify Base and Height

Diagram 1

From the figure: Base (AD)=8 cm\text{Base } (AD) = 8\text{ cm} Height=10 cm\text{Height} = 10\text{ cm}

Step 2 · Calculate Area of Triangle ADE

Area of triangle ADE=12×base×height=12×8 cm×10 cm=4 cm×10 cm=40 cm2\begin{aligned} \text{Area of triangle } ADE &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times 8\text{ cm} \times 10\text{ cm} \\[0.6em] &= 4\text{ cm} \times 10\text{ cm} \\[0.6em] &= 40\text{ cm}^2 \end{aligned}
Answer

40 cm240\text{ cm}^2

Common Mistakes
  • Using Slant Lengths as Height: Mistaking the slanted side lengths AEAE or DEDE for the height instead of using the perpendicular height (10 cm10\text{ cm}).
  • Omitting the 12\dfrac{1}{2} Factor: Calculating base×height=80 cm2\text{base} \times \text{height} = 80\text{ cm}^2, which is the area of the enclosing rectangle rather than the triangle.

More questions in Exercise 6.2

Q1

Find the area of triangle ADE in Fig. 6.31.

Q2

The parallel sides of a trapezium are 40 cm40\text{ cm} and 20 cm20\text{ cm}. If its non-parallel sides are both equal, each being 26 cm26\text{ cm}, find the area of the trapezium.

Q3

Find the area of a triangle, given that its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm} long, and its perimeter is 32 cm32\text{ cm}.

Q4

The sides of a triangular plot are in the ratio 3:5:73 : 5 : 7; its perimeter is 300 m300\text{ m}. Find its area.

Q5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2128\text{ cm}^2, find the length of the shorter diagonal.

Q6

ABCDABCD is a parallelogram. PP and QQ are any two points on side ABAB. What can you say about the ratio area(ΔPCD):area(ΔQCD)\text{area}(\Delta \text{PCD}) : \text{area}(\Delta \text{QCD})?

Q7

OO is any point on the diagonal PRPR of a parallelogram PQRSPQRS. Prove that the areas of triangles PSOPSO and PQOPQO are equal.

Q8

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Q9

In ΔABC\Delta\text{ABC}, the midpoint of BC\text{BC} is D\text{D} (Fig. 6.32). Median AD\text{AD} is drawn. P\text{P} is any point on AD\text{AD}. Show that area(ΔABP)=area(ΔACP)\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP}).

Q10

Given a square ABCDABCD, let PP be a point within it. Join PAPA, PBPB, PCPC, PDPD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB\Delta PAB and ΔPCD\Delta PCD) and the green region (ΔPBC\Delta PBC and ΔPDA\Delta PDA)?

Q11

In ΔABC\Delta\text{ABC}, DD is the midpoint of AB\text{AB}. PP is any point on BC\text{BC}, and QQ is a point on AB\text{AB} such that CQPD\text{CQ} \parallel \text{PD}. PQ\text{PQ} is joined (Fig. 6.34). Prove that

Area(ΔBPQ)=12Area(ΔABC)\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})

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