Q1. Find the area of triangle ADE in Fig. 6.31. (10 cm, 8 cm, E, A, D, B, C)

Answer:

From Fig. 6.31, we see triangle ADE is inside rectangle ABCD. Given AB = 10 cm, BC = 8 cm. Since ABCD is a rectangle, AD = BC = 8 cm and AB = CD = 10 cm. Triangle ADE has base AD = 8 cm and height from E to AD equals AB = 10 cm. Area = (1/2) × base × height = (1/2) × 8 × 10 = 40 cm².

Q2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

Answer:

Let the trapezium be ABCD with AB ∥ CD, AB = 40 cm, CD = 20 cm, AD = BC = 26 cm. Draw perpendiculars from C and D to AB meeting at E and F. Let AE = FB = x. Then EF = CD = 20 cm. So AB = AE + EF + FB = x + 20 + x = 40, so 2x = 20, x = 10 cm. Height h = √(AD² - AE²) = √(26² - 10²) = √(676 - 100) = √576 = 24 cm. Area = (1/2) × (sum of parallel sides) × height = (1/2) × (40 + 20) × 24 = (1/2) × 60 × 24 = 720 cm².

Q3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Answer:

Let the sides be a = 8 cm, b = 11 cm, c = unknown. Perimeter = a + b + c = 32, so c = 32 - 8 - 11 = 13 cm. Semi-perimeter s = 32/2 = 16 cm. Using Heron's formula: Area = √[s(s-a)(s-b)(s-c)] = √[16(16-8)(16-11)(16-13)] = √[16 × 8 × 5 × 3] = √(1920) = √(64 × 30) = 8√30 ≈ 8 × 5.477 = 43.82 cm².

Q4. The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

Answer:

Let sides be 3k, 5k, 7k. Perimeter = 3k + 5k + 7k = 15k = 300, so k = 20. Sides: a = 60 m, b = 100 m, c = 140 m. Semi-perimeter s = 300/2 = 150 m. Area = √[150(150-60)(150-100)(150-140)] = √[150 × 90 × 50 × 10] = √[150 × 90 × 500] = √(6,750,000) = √(25 × 27 × 10000) = 100 × 5 × √27 = 500 × 3√3 = 1500√3 ≈ 2598 m².

Q5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

Answer:

Let diagonals be d1 and d2 with d2 = 2d1. Area of rhombus = (1/2) × d1 × d2 = (1/2) × d1 × 2d1 = d1² = 128. So d1 = √128 = 8√2 cm. The shorter diagonal is d1 = 8√2 cm.

Q6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (△PCD) : area (△QCD)?

Answer:

In parallelogram ABCD, triangles PCD and QCD share the same base CD. Their heights are the perpendicular distances from P and Q to CD. Since P and Q lie on AB, and AB ∥ CD, the perpendicular distance from any point on AB to CD is constant (equal to the height of the parallelogram). Therefore, both triangles have equal area. So the ratio area(△PCD) : area(△QCD) = 1 : 1.

Q7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Answer:

In parallelogram PQRS, diagonal PR divides it into two equal triangles: △PQR and △PSR. O lies on PR. Triangles PSO and PQO share the same base PO. Their heights are distances from S and Q to line PR. Since PR is a diagonal, S and Q are symmetric with respect to PR (the parallelogram is symmetric about its diagonal), so distances are equal. Thus, area(△PSO) = area(△PQO). Alternatively, since △PSR and △PQR have equal area, and triangles OSR and OQR share base OR and have equal heights (from S and Q to PR), area(△OSR) = area(△OQR). So area(△PSO) = area(△PSR) - area(△OSR) = area(△PQR) - area(△OQR) = area(△PQO).

Q8. If the mid-points of the sides of a 4-gon (quadrilateral) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

Answer:

Let ABCD be the 4-gon. Let P, Q, R, S be midpoints of AB, BC, CD, DA respectively. Join PQRS. By the midpoint theorem, PQ ∥ AC and SR ∥ AC, so PQ ∥ SR. Similarly, PS ∥ BD and QR ∥ BD, so PS ∥ QR. Hence PQRS is a parallelogram. Join diagonals AC and BD. The area of parallelogram PQRS = (1/2) × area of quadrilateral formed by joining midpoints (Varignon's theorem). Specifically, area(PQRS) = (1/2) × (area of quadrilateral formed by joining the midpoints of the diagonals). More directly, area(PQRS) = (1/2) × area(ABCD). This can be proved by showing that the sum of areas of four corner triangles equals the area of the parallelogram.

Q9. In △ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (△ABP) = area (△ACP).

Answer:

Since D is midpoint of BC, AD is median. Triangles ABD and ACD have equal areas (since equal bases BD = DC and same height from A). P lies on AD. Triangles PBD and PCD have equal areas (same base BD = DC and same height from P to BC). Now, area(△ABP) = area(△ABD) - area(△PBD) = area(△ACD) - area(△PCD) = area(△ACP). Therefore, area(△ABP) = area(△ACP).

Q10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (△PAB and △PCD) and the green region (△PBC and △PDA)?

Answer:

Let the side of the square be s. Draw perpendiculars from P to the sides. Let the distances from P to AB, BC, CD, DA be h1, h2, h3, h4 respectively. Then h1 + h3 = s and h2 + h4 = s. Area(△PAB) = (1/2) × s × h1, area(△PCD) = (1/2) × s × h3, so sum of red areas = (1/2)s(h1 + h3) = (1/2)s². Similarly, sum of green areas = (1/2)s(h2 + h4) = (1/2)s². Therefore, the ratio of red to green is 1 : 1.

Q11. In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined (Fig. 6.34). Prove that Area (△BPQ) = 1/2 Area (△ABC).

Answer:

Since D is midpoint of AB, AD = DB. CQ ∥ PD. In △ABP, PD ∥ CQ and D is midpoint of AB, so by converse of midpoint theorem, Q is midpoint of AB? Actually, since PD ∥ CQ and D is midpoint of AB, line through D parallel to CQ meets AB at Q such that DQ = ? Consider triangles: PD ∥ CQ implies that in △BCQ, PD is a line through the midpoint of BQ? Better: Join CD. Area(△BDC) = (1/2) area(△ABC) (since D is midpoint). Since PD ∥ CQ, triangles PDQ and CDQ have equal areas (same base DQ and same height between parallels). Also, area(△BPQ) = area(△BPD) + area(△PDQ). And area(△BCD) = area(△BPD) + area(△PDC). Since area(△PDQ) = area(△CDQ), and area(△PDC) = area(△CDQ) + area(△CQP)? This requires careful reasoning. Alternatively, use that area(△BPQ)/area(△ABC) = (BP × BQ)/(BC × BA). Using similar triangles, BQ/BA = BP/BC, so area ratio = (BP/BC)². Since P is arbitrary, this doesn't give 1/2 unless P is specific. The problem statement may have an error or require additional condition. However, if Q is such that CQ ∥ PD and D is midpoint, then by properties of similar triangles, BQ = (1/2)BA and BP = (1/2)BC, making P the midpoint of BC. Then area(△BPQ) = (1/2)×(BC/2)×(AB/2)×sinB = (1/4)×(1/2)×BC×AB×sinB = (1/4) area(△ABC). This contradicts the claim. The correct proof: Since CQ ∥ PD, triangles PDQ and CDQ have same area. Also, D is midpoint, so area(△BCD) = (1/2) area(△ABC). Now, area(△BPQ) = area(△BPD) + area(△PDQ) = area(△BPD) + area(△CDQ) = area(△BCD) = (1/2) area(△ABC). This works if Q lies between D and B such that DQ = QB? Actually, careful: area(△BPD) + area(△CDQ) = area(△BCD) only if points align properly. Since PD ∥ CQ, we have PD and CQ as parallel lines, and triangles BPD and BCD share vertex B but have different bases. The standard result: When D is midpoint of AB and CQ ∥ PD, then Q is the midpoint of AB? No, Q is on AB, PD ∥ CQ, so in triangle ABP, line through D parallel to CQ meets AP at some point. Actually, the correct geometric property: area(△BPQ) = area(△BPD) + area(△PDQ). Since PD ∥ CQ, area(△PDQ) = area(△CDQ). Also, area(△BCD) = (1/2) area(△ABC). Now, area(△BPD) + area(△CDQ) = area(△BCD) - area(△PCQ)? This is messy. The intended solution uses that triangles BPQ and ABC share angle B, and using the midline theorem, Q is the midpoint of AB, making area(△BPQ) = (1/2) area(△ABP) = (1/4) area(△ABC) — still not 1/2. Possibly the problem meant area(△BPQ) = area(△ABC)/2 when P is the midpoint of BC. Without additional conditions, the statement as given may be incomplete.