Measuring Space: Perimeter and Area | Exercise 6.2

Question 6

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})?

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Solution
Understand the Question
  • Two triangles that share the same base and lie between the same parallel lines have equal altitudes (heights) and therefore equal areas.
  • In parallelogram ABCDABCD, side ABCDAB \parallel CD. Since both points PP and QQ lie on line ABAB, triangles ΔPCD\Delta \text{PCD} and ΔQCD\Delta \text{QCD} share the base CDCD and have the same perpendicular height from line ABAB to line CDCD.
  • Therefore, their areas are equal, making the ratio of their areas 1:11 : 1.

Step 1 · Identify Common Base and Height

Both triangles ΔPCD\Delta \text{PCD} and ΔQCD\Delta \text{QCD} lie on the common base CDCD.Diagram 1

Since ABCDABCD is a parallelogram, ABCDAB \parallel CD.

Points PP and QQ both lie on line ABAB. The perpendicular distance between two parallel lines is constant everywhere, so the height hh from PP to base CDCD is equal to the height from QQ to base CDCD.

Step 2 · Compare Areas and Calculate the Ratio

Using the area formula for a triangle Area=12×base×height\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}

Area(ΔPCD)=12×CD×h\text{Area}(\Delta \text{PCD}) = \dfrac{1}{2} \times CD \times h

Area(ΔQCD)=12×CD×h\text{Area}(\Delta \text{QCD}) = \dfrac{1}{2} \times CD \times h

Since base and height are identical Area(ΔPCD)=Area(ΔQCD)\text{Area}(\Delta \text{PCD}) = \text{Area}(\Delta \text{QCD})

Therefore, the ratio of their areas is area(ΔPCD):area(ΔQCD)=1:1\text{area}(\Delta \text{PCD}) : \text{area}(\Delta \text{QCD}) = 1 : 1

Answer

1:11 : 1

Common Mistakes
  • Assuming Position Affects Area: Mistakenly believing that because PP and QQ are at different locations along ABAB, the triangles have different areas. The perpendicular distance from any point on ABAB to line CDCD is constant.
  • Confusing Slant Length with Height: Confusing side lengths like PDPD or QDQD with the perpendicular altitude hh of the triangles.

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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