Exercise 72
27: Find the reas
(2) D4-5
1: a Puty=2in10
weg 10,-
Puty =21
1. Add:
(a) 9x + y - 3z and -4X-5y +7z.
(b) m – 7m2 – 6m+3,-2m2 +8m² +6m–12 and 13–9m+m+5m2.
(c) 5 pgr, -9pgr, 7pgr and -4 pgr.
(d) -x+11y2 – 3xy and x3 – 17x2 – 10y? +9xy.
(e) 8z? – 2+2, 117–322 -5 and 922–4 +7z.
(f) 60+5b and 2a-8b.
2. Subtract:
(a) 2p-3g from 7p+q.
(b) x2 + 8xy – 3y2 from 5x2 – 9y2.
(c) -62 – 40+14 from 60 -13.
(d) 24 – y + 3z²y2 from 42€ - 5y?z? +7y4.
(e) 20° +302 +40-5 from -403-80+1202 +15.
(f) P-1 from 913 – 712-4.
312
603_502.30 Find
Weget
(c) Puty
ple 8: Fid
ande
-
Answers
Answered by
1
Answer:
swer
To prove:
A(2,−2),B(−2,1) and C(5,2) are the vertices of a right angled-triangle.
Proof:
By distance formula,
AB=
(−2−2)
2
+(1+2)
2
=
16+9
=
25
=5
BC=
(−2−5)
2
+(1−2)
2
=
(−7)
2
+(−1)
2
=
50
=5
2
Length of the hypotenuse
AC=
(5−2)
2
+(2+2)
2
=
(3)
2
+(4)
2
=
25
∴AB
2
+AC
2
=25+25=50=BC
2
∴ △ ABC is a right-angled triangle
Area of the triangle ABC
=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]sq.
Area of the △ABC
=
2
1
×Base×height
=
2
1
×5×5
=
2
25
sq.units.
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