Advertisements
Advertisements
प्रश्न
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°, PA = 6 cm and QA = 8 cm. Find ∠PQR
विकल्प
80°
85°
75°
90°
उत्तर
90°
Explanation;
Hint:
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°
In the ∆PQR,
PQ = `sqrt("PR"^2 - "QR"^2)`
= `sqrt(26^2 - 24^2)`
= `sqrt(676 - 576)`
= `sqrt(100)`
= 10
In the right ∆APQ
PQ2 = PA2 + AQ2
= 62 + 82
= 36 + 64
= 100
PQ = `sqrt(100)`
= 10
∆PQR is a right angled triangle at Q.
Since PR2 = PQ2 + QR2
∠PQR = 90°
APPEARS IN
संबंधित प्रश्न
If ΔABC ~ ΔPQR and ∠A = 60°, then ∠P = ?
In the following figure, point D divides AB in the ratio 3 : 5. Find :
DE = 2.4 cm, find the length of BC.
In ΔPQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP.
Prove that : (i) ΔPQL ∼ ΔRPM
(ii) QL. Rm = PL. PM
(iii) PQ2 = QR. QL.
In the adjoining figure, the medians BD and CE of a ∆ABC meet at G. Prove that
(i) ∆EGD ∼ ∆CGB and
(ii) BG = 2GD for (i) above.
In ΔABC, point D divides AB in the ratio 5:7, Find: DE, If BC = 4.8cm
In ΔABC, DE is drawn parallel to BC cutting AB in the ratio 2 : 3. Calculate:
(i) `("area"(Δ"ADE"))/("area"(Δ"ABC")`
(i) `("area"("trapeziumEDBC"))/("area"(Δ"ABC"))`
The perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, then the length of AB is
Areas of two similar triangles are 225 cm2 and 81 cm2. If side of smaller triangle is 12 cm, find corresponding side of major triangle.
∆ABC ~ ∆PQR. If AM and PN are altitudes of ΔABC and ∆PQR respectively and AB2 : PQ2 = 4 : 9, then AM : PN = ______.
It is given that ΔABC ~ ΔPQR, with `(BC)/(QR) = 1/3`. Then, `(ar(PRQ))/(ar(BCA))` is equal to ______.
