Measuring Space: Perimeter and Area | Exercise 6.2

Question 4

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.

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Solution
Understand the Question
  • The side lengths of the triangle are in the ratio 3:5:73 : 5 : 7. We represent them as 3x3x, 5x5x, and 7x7x.
  • Using the given perimeter of 300 m300\text{ m}, we solve for xx to find the actual lengths of all three sides.
  • Once the sides are known, we compute the semi-perimeter s=Perimeter2s = \dfrac{\text{Perimeter}}{2} and apply Heron's formula to find the area: Area=s(sa)(sb)(sc)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

Step 1 · Find the Side Lengths

Let the sides of the triangular plot be 3x3x, 5x5x, and 7x7x.Diagram 1

Given that the perimeter is 300 m300\text{ m}

3x+5x+7x=30015x=300x=30015=20\begin{aligned} 3x + 5x + 7x &= 300 \\ 15x &= 300 \\[0.6em] x &= \dfrac{300}{15} = 20 \end{aligned}

Calculating each side length: a=3×20=60 ma = 3 \times 20 = 60\text{ m} b=5×20=100 mb = 5 \times 20 = 100\text{ m} c=7×20=140 mc = 7 \times 20 = 140\text{ m}

Step 2 · Calculate the Semi-Perimeter

Semi-perimeter ss is given by

s=Perimeter2=3002=150 m\begin{aligned} s &= \dfrac{\text{Perimeter}}{2} \\[0.6em] &= \dfrac{300}{2} = 150\text{ m} \end{aligned}

Step 3 · Calculate Area Using Heron's Formula

By Heron's formula Area=s(sa)(sb)(sc)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

Substitute s=150 ms = 150\text{ m}, a=60 ma = 60\text{ m}, b=100 mb = 100\text{ m}, and c=140 mc = 140\text{ m}

Area=150(15060)(150100)(150140)=150×90×50×10=6750000=225×3×10000=225×3×10000=15×3×100=15003 m2\begin{aligned} \text{Area} &= \sqrt{150(150-60)(150-100)(150-140)} \\ &= \sqrt{150 \times 90 \times 50 \times 10} \\ &= \sqrt{6750000} \\ &= \sqrt{225 \times 3 \times 10000} \\ &= \sqrt{225} \times \sqrt{3} \times \sqrt{10000} \\ &= 15 \times \sqrt{3} \times 100 \\ &= 1500\sqrt{3}\text{ m}^2 \end{aligned}
Answer

15003 m21500\sqrt{3}\text{ m}^2

Common Mistakes
  • Ratio Misinterpretation: Directly substituting the ratio values (3,5,73, 5, 7) into Heron's formula instead of finding the actual side lengths (60 m,100 m,140 m60\text{ m}, 100\text{ m}, 140\text{ m}).
  • Using Perimeter instead of Semi-Perimeter: Substituting the perimeter s=300 ms = 300\text{ m} rather than the semi-perimeter s=150 ms = 150\text{ m} into Heron's formula.
  • Calculation Errors under Square Root: Multiplying into very large numbers without factoring perfect squares (225225 and 1000010000), which complicates finding the square root.

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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