(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.
(b)
In the diagram \(\overline{PQ} || \overline{YZ}\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \(\Delta\) XPQ = 24\(cm^{2}\).Calculate the area of the trapezium PQZY.
(b)
In the diagram \(\overline{PQ} || \overline{YZ}\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \(\Delta\) XPQ = 24\(cm^{2}\).Calculate the area of the trapezium PQZY.
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Correct Answer: Option n
Explanation:
(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P)
3rd term = -9 ; 8th term = 26
\(\implies T_{3} = a + 2d = -9 ... (1)\)
\(\implies T_{8} = a + 7d = 26 ..... (2)\)
(i) Solving (2) - (1), we have
\(7d - 2d = 26 - (-9)\)
\(5d = 35 \implies d = 7\)
(ii) Putting d = 7 in (1) above, we have
\(a + 2(7) = -9\)
\(a + 14 = -9\)
\(a = - 9 - 14 = -23\)
(b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles)
\(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\)
\(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\)
\(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)
(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P)
3rd term = -9 ; 8th term = 26
\(\implies T_{3} = a + 2d = -9 ... (1)\)
\(\implies T_{8} = a + 7d = 26 ..... (2)\)
(i) Solving (2) - (1), we have
\(7d - 2d = 26 - (-9)\)
\(5d = 35 \implies d = 7\)
(ii) Putting d = 7 in (1) above, we have
\(a + 2(7) = -9\)
\(a + 14 = -9\)
\(a = - 9 - 14 = -23\)
(b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles)
\(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\)
\(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\)
\(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)