please tell me the answer with full explanation
Answers
Answer:
18 question k answer
in r. t. abc
using p. t.
ab²= ac²+bc²
bc²= 25²-7²
bc = 625-49
bc=24 CM
Answer:
Question 16
Steps for construction:-
1).Draw a line segment AB of 5cm.
2).Taking A and B as centre, draw arcs of 7cm and 6cm radius respectively.
3).Let these arcs intersect each other at point C.
Join AB and AC.
4).△ABC is the required triangle having length of sides as 5cm,7cm and 6cm respectively.
Question 17
PQR +QRP+RPQ=180
105+ 40+ RPQ=180
145+ RPQ=180
RPQ= 180-145
RPQ= 35
LET DRAW A LINE PQ = 5CM
DRAW ANGLE P = 35°
DRAW ANGLE Q = 105°
FINALLY THEY INTERSECT EACH OTHER AT POINT R= 40°
Question 18
By applying Pythagoras theorem in ∆ABC
(AC)² + (BC)² = (AB)²
(BC)² = (AB)² – (AC)²
(BC)² = (25)² – (7)²
(BC)² = 625 – 49 = 576
BC = 24cm
or
Let A'CB represents the tree before it broken at the point C and let the top A' touches the ground at A after it broke. Then ΔABC is a right angled triangle, right angled at B.
AB=12m and BC=5m
Using Pythagoras theorem, In ΔABC
(AC)² + (AB)² + (BC)²
⇒(AC)² = (12)² + (5)²
⇒(AC)² = 144 + 25
⇒(AC)² = 169
⇒AC=13m
Hence, the total height of the tree=AC+CB=13+5=18m.
Question 19
To construct: An isosceles triangle PQR whrere PQ=RQ=6.5cm and ∠Q=110°
Steps of construction:
(a) Draw a line segment QR=6.5cm
(b) At point Q, draw an angle of 110°
with the help of protractor, i.e., ∠YQR=110°
(c) Taking Q as centre, draw an arc with radius 6.5cm which cuts QY at point P
(d) Join PR
It is the required isosceles triangle PQR