Question

С
E
B
(a)
11. In AABC, ZABC = 90° and BD I AC. If AD = 2 cm and BD = 4 cm, find the following.
ar(ABD) ar(ABCD)
(b)
ar(AABC)
(c)
ar(ABCD)
ar(4ABC) ar(AABD)
12. In APQR, ZPQR = 90°.QS 1 PR and ST 1 QR. If PQ = 6 cm and PS = 3 cm, find the following.
(a)
ar(AQST)
ar(AQST)
ar(APQS)
ar(ARST)
ar(AQRS)
(c)
ar(AQST)
ar(AQST)
ar(APQR)
ar(ARST) ar(trapezium PQTS)
ar(APQR)
ar(APQR)
(b)
P
3 cm
S
(d)
6 cm
Q
R
M
13. AXYZ is an isosceles triangle in which XY = XZ = 13 cm and
YZ = 10 cm. XN 1 YZ. If ZW = 8 cm and WM 1 XY, find the
following
ar(AXNZ)
(a)
ar(AMWY)
ar(AMWY)
ar(AXYZ)
(b)
w
N
Z
D
С
P
(a)
А
la
B
R
14. ABCD is a rhombus with side equal to 8 cm. CB is produced to R
such that BR =4 cm. DR cuts AC and AB at P and Q respectively.
Find
ar(AADP)
ar(ADCP)
(b)
ar(APCR)
ar(AAPQ)
ar(AADP)
(c)
ar(AADP)
(d)
ar(ADCP)
ar(AAPQ)
15. In AABC, DE is drawn parallel to BC such that
AD = 35 cm, DB = 65 cm and DC intersects BE at
F. Find
ar(ADEF)
ar(ADEF)
(a)
(b)
ar(ABCF)
ar(AECF)
ar(ADEF)
ar(AADE)
(c)
(d)
ar(ADEB)
ar(trapezium DBCE)
А
D
E
F
B
с
A
F
P.
E
16. In AABC, D, E and Fare the midpoints of BC, CA
and AB respectively. Again, P, Q and R are the
midpoints of EF, FD and DE respectively. Find
ar(AABC)
ar(APQR)
(a)
(b)
ar(ABDF)
ar(AABC)
ar(APQR)
(c)
ar(AEDC)
R
Q
B
С
D​

Answer 1

Answer:

Step-by-step explanation:

In quadrilateral ABCD we have                  AC = AD            and AB being the bisector of ∠A.            Now, in ΔABC and ΔABD,                  AC = AD[Given]                  AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD]            ∴ Using SAS criteria, we have                  ΔABC ≌ ΔABD.            ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.            ∴ BC = BD.