Area of a Isoceles Triangle calculator

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Area of a Isoceles Triangle

Height (h)=√a2-b24 $(h) = sqrt(a^2 - b^2/4)$
Perimeter (P)=2a+b $(P) = 2 a + b$
Area (A)=bh2 $(A) = (b h)/2$

Example :
I know that for a isoceles triangles a = 5 and b = 6 . From this find out Area of the isoceles triangles.

Here base (b)=6 and equal side (a)=5 (Given) $"Here base " (b) = 6" and equal side " (a) = 5" (Given)"$

We know that, $"We know that,"$

Perimeter=(2⋅equal Side)+third Side $"Perimeter" = (2 * "equal Side") + "third Side"$

=(2⋅a)+b $= (2 * a) + b$

=(2⋅5)+6 $= (2 * 5) + 6$

=16 $= 16$

We know that, $"We know that,"$

Area=12⋅base⋅height $"Area" = 1/2 * "base" * "height"$

=12⋅b⋅√a2-b24 $= 1/2 * b * sqrt(a^2 - b^2/4)$

=12⋅6⋅√52-624 $= 1/2 * 6 * sqrt(5^2 - 6^2/4)$

=12⋅6⋅4 $= 1/2 * 6 * 4$

=12 $= 12$

Area

1. Circle

Area

(A)=πr2 $(A) = pi r^2$
Circumference

(C)=2πr=πd $(C) = 2 pi r = pi d$
Diameter

(d)=2r $(d) = 2 r$

2. Semicircle

Area

(A)=12πr2 $(A) = 1/2 pi r^2$
Circumference

(C)=πr=πd2 $(C) = pi r = (pi d)/2$
Perimeter

(P)=πr+2r $(P) = pi r + 2 r$
Diameter

(d)=2r $(d) = 2 r$

3. Regular Hexagon

Perimeter

(P)=6a $(P) = 6 a$
Area

(A)=√34×6×a2 $(A) = sqrt(3)/4 xx 6 xx a^2$

4. Square

Diagonal

(d)=√2a $(d) = sqrt(2) a$
Perimeter

(P)=4a $(P) = 4a$
Area

(A)=a2=d22 $(A) = a^2 = d^2/2$

5. Rectangle

Diagonal

(d)=√l2+b2 $(d) = sqrt(l^2 + b^2)$
Perimeter

(P)=2(l+b) $(P) = 2(l+b)$
Area

(A)=lb $(A) = l b$

6. Parallelogram

Area

(A)=ah $(A) = ah$
Perimeter

(P)=2a+2b $(P) = 2a + 2b$

7. Rhombus

Radius

(r1)=d12 $(r_1) = (d_1)/2$
Radius

(r2)=d22 $(r_2) = (d_2)/2$
Side

(a)=√r21+r22 $(a) = sqrt(r_1^2 + r_2^2)$
Perimeter

(P)=4a $(P) = 4 a$
Area

(SA)=d1d22 $(SA) = (d_1 d_2)/2$

8. Trapezium

Area

(A)=h2(a+b) $(A) = h/2 (a + b)$
Perimeter

(P)=a+b+c+d $(P) = a + b + c + d$

Area >> Triangle

9. Scalene Triangle

Perimeter

(P)=a+b+c $(P) = a+b+c$

S=P2=a+b+c2 $S = P/2 = (a+b+c)/2$
Area

(A)=√S(S-a)(S-b)(S-c) $(A) = sqrt(S (S - a) (S - b) (S - c))$

10. Right angle Triangle

Diagonal

(d)=√a2+b2 $(d) = sqrt(a^2 + b^2)$
Perimeter

(P)=a+b+c $(P) = a+b+c$
Area

(A)=12(ab) $(A) = 1/2(a b)$

11. Equilateral Triangle

Perimeter

(P)=3a $(P) = 3 a$
Area

(A)=√34a2 $(A) = sqrt(3)/4 a^2$

12. Isosceles Triangle

Height

(h)=√a2-b24 $(h) = sqrt(a^2 - b^2/4)$
Perimeter

(P)=2a+b $(P) = 2 a + b$
Area

(A)=bh2 $(A) = (b h)/2$

13. Sector Segment

Length of the arc

=l=πrθ180 $= l = (pi r theta)/180$
Area of a minor sector

=πr2θ360 $= (pi r^2 theta)/360$

Problem : 1 / 13 [ Circle ] Enter your problem 1. I know that for a circle Radius = 10 . From this find out Area of the circle.


Solution:	Here radius (r)=10(Given) $"Here radius " (r)=10 "(Given)"$ Diameter (d)=2⋅r $"Diameter " (d)= 2 * r$ =2⋅10 $= 2 * 10$ =20 $= 20$ Perimeter=2πr $"Perimeter" = 2 pi r$ =2⋅227⋅10 $= 2 * 22/7 * 10$ =4407 $= 440/7$ Area=πr2 $Area = pi r^2$ =227⋅(10)2 $= 22/7 * (10)^2$ =22007 $= 2200/7$

Problem : 2 / 13 [ Semicircle ] Enter your problem 2. I know that for a Semicircle Radius = 10 . From this find out Area of the Semicircle.


Solution:	Here radius (r)=10 (Given) $"Here radius "(r)=10" (Given)"$ Diameter (d)=2r $"Diameter "(d)= 2 r$ =2⋅10 $= 2 * 10$ =20 $= 20$ Circumference=πr $"Circumference" = pi r$ =227⋅10 $= 22/7 * 10$ =2207 $= 220/7$ Perimeter=πr+2r $"Perimeter" = pi r + 2 r$ =227⋅10+2⋅10 $= 22/7 * 10 + 2 * 10$ =2207+20 $= 220/7 + 20$ =3607 $= 360/7$ Area=πr22 $"Area" = (pi r^2)/2$ =227⋅(10)22 $= (22/7 * (10)^2)/2$ =11007 $= 1100/7$

Problem : 3 / 13 [ RegularHexagon ] Enter your problem 3. I know that for a Regular Hexagon Side = 10 . From this find out Area of the Regular Hexagon.


Solution:	Here Side (a)=10 (Given) $"Here Side "(a)=10" (Given)"$ Perimeter=6a $"Perimeter" = 6 a$ =6⋅10 $= 6 * 10$ =60 $= 60$ Area=√34⋅6⋅a2 $"Area" = sqrt(3)/4 * 6 * a^2$ =√34⋅6⋅102 $= sqrt(3)/4 * 6 * 10^2$ =259.8076 $= 259.8076$

Problem : 4 / 13 [ Square ] Enter your problem 4. I know that for a square Side(a) = 10 . From this find out Area of the square.


Solution:	Here a=10 (Given) $"Here a" = 10" (Given)"$ Diagonal=√2⋅a $"Diagonal" = sqrt(2) * "a"$ =√2⋅10 $= sqrt(2) * 10$ =14.1421 $= 14.1421$ Perimeter=4⋅a $"Perimeter" = 4 * "a"$ =4⋅10 $= 4 * 10$ =40 $= 40$ Area=a2 $"Area" = "a"^2$ =102 $= 10^2$ =100 $= 100$

Problem : 5 / 13 [ Rectangle ] Enter your problem 5. I know that for a rectangle Length = 5 and Breadth = 12 . From this find out Area of the rectangle.


Solution:	Here one-Side (l)=5 and other-Side (b)=12 (Given) $"Here one-Side " (l) = 5" and other-Side " (b) = 12" (Given)"$ Diagonal=√l2+b2 $"Diagonal" = sqrt(l^2 + b^2)$ =√52+122 $= sqrt(5^2 + 12^2)$ =√25+144 $= sqrt(25 + 144)$ =√169 $= sqrt(169)$ =13 $= 13$ Perimeter=2⋅(sum of Sides) $"Perimeter" = 2 * "(sum of Sides)"$ =2⋅(5+12) $= 2 * (5 + 12)$ =34 $= 34$ Area = Product of Sides $"Area = Product of Sides"$ =5⋅12 $= 5 * 12$ =60 $= 60$

Problem : 6 / 13 [ Parallelogram ] Enter your problem 6. I know that for a parallelogram a = 9 , b = 22 and h = 14 . From this find out Area of the parallelogram.


Solution:	Here a=9,b=22,h=14 (Given) $"Here " a=9, b=22, h=14" (Given)"$ Perimeter=2⋅(a+b) $"Perimeter" = 2 * (a + b)$ =2⋅(9+22) $= 2 * (9 + 22)$ =62 $= 62$ Area=a⋅h $"Area" = a * h$ =9⋅14 $= 9 * 14$ =126 $= 126$

Problem : 7 / 13 [ Rhombus ] Enter your problem 7. I know that for a rhombus d1 = 10 and d2 = 24 . From this find out Area of the rhombus.


Solution:	Here, we have d1=10 and d2=24 (Given) $"Here, we have " d_1 = 10" and " d_2 = 24" (Given)"$ a2=(d12)2+(d22)2 $a^2 = (d_1/2)^2 + (d_2/2)^2$ a2=(102)2+(242)2 $a^2 = (10/2)^2 + (24/2)^2$ a2=(5)2+(12)2 $a^2 = (5)^2 + (12)^2$ a2=169 $a^2 = 169$ a=13 $a = 13$ Perimeter=4⋅a $"Perimeter" = 4 * a$ =4⋅13 $= 4 * 13$ =52 $= 52$ Area=12 (Product of diagonals) $"Area" = 1/2 " (Product of diagonals)"$ =12⋅d1⋅d2 $= 1/2 * d_1 * d_2$ =12⋅10⋅24 $= 1/2 * 10 * 24$ =120 $= 120$

Problem : 8 / 13 [ Trapezium ] Enter your problem 8. I know that for a trapezium a = 22 , b = 18 , c = 16 , d = 16 and h = 4 . From this find out Area of the trapezium.


Solution:	Here a=22,b=18,c=16,d=16,h=4 (Given) $"Here " a=22, b=18, c=16, d=16, h=4" (Given)"$ Perimeter=a+b+c+d $"Perimeter" = a + b + c + d$ =22+18+16+16 $= 22 + 18 + 16 + 16$ =72 $= 72$ Area=(a+b)⋅h2 $"Area" = (a + b) * h/2$ =(22+18)⋅42 $= (22 + 18) * 4/2$ =80 $= 80$

Problem : 9 / 13 [ Scalene Triangle ] Enter your problem 9. I know that for a scalene triangles a = 3 , b = 4 and c = 5 . From this find out Area of the scalene triangles.


Solution:	Here a=3,b=4,c=5 (Given) $"Here " a=3, b=4, c=5" (Given)"$ We know that, $"We know that,"$ Perimeter=a+b+c $"Perimeter" = a + b + c$ =3+4+5 $= 3 + 4 + 5$ =12 $= 12$ Semi-Perimeter=s=a+b+c2 $"Semi-Perimeter" = s = (a + b + c)/2$ =122 $= 12/2$ =6 $= 6$ Here a=3,b=4,c=5 and semi-Perimeter=6 $"Here " a=3, b=4, c=5" and semi-Perimeter" = 6$ We know that, $"We know that,"$ Area=√s(s-a)(s-b)(s-c) $"Area" = sqrt(s (s - a) (s - b) (s - c))$ =√6(6-3)(6-4)(6-5) $= sqrt(6 (6 - 3) (6 - 4) (6 - 5))$ =6 $= 6$

Problem : 10 / 13 [ Rightangle Triangle ] Enter your problem 10. I know that for a rightangle triangles AB = 5 and BC = 12 . From this find out Area of the rightangle triangles.


Solution:	Here one Side=5 and other Side=12 (Given) $"Here one Side" = 5" and other Side" = 12" (Given)"$ We know that, $"We know that,"$ In triangle ABC, by Pythagoras' theorem $"In triangle ABC, by Pythagoras' theorem"$ AC2=AB2+BC2 $AC^2 = AB^2 + BC^2$ AC2=52+122 $AC^2 = 5^2 + 12^2$ AC2=25+144 $AC^2 = 25 + 144$ AC2=169 $AC^2 = 169$ AC=13 $AC = 13$ Perimeter=AB+BC+AC $"Perimeter" = AB + BC + AC$ =5+12+13 $= 5 + 12 + 13$ =30 $= 30$ Here base=5 and height = 12 $"Here base" = 5" and height = "12$ We know that, $"We know that,"$ Area=12⋅AB⋅BC $"Area" = 1/2 * AB * BC$ =12⋅5⋅12 $= 1/2 * 5 * 12$ =30 $= 30$

Problem : 11 / 13 [ Equilateral Triangle ] Enter your problem 11. I know that for a equilateral triangles Side = 6 . From this find out Area of the equilateral triangles.


Solution:	Here a=6 (Given) $"Here " a = 6" (Given)"$ We know that, $"We know that,"$ Perimeter=3⋅a $"Perimeter" = 3 * a$ =3⋅6 $= 3 * 6$ =18 $= 18$ We know that, $"We know that,"$ Area=√34⋅a2 $"Area" = sqrt(3)/4 * a^2$ =1.7324⋅6⋅6 $= 1.732/4 * 6 * 6$ =15.5885 $= 15.5885$

Problem : 12 / 13 [ Isoceles Triangle ] Enter your problem 12. I know that for a isoceles triangles a = 5 and b = 6 . From this find out Area of the isoceles triangles.


Solution:	Here base (b)=6 and equal side (a)=5 (Given) $"Here base " (b) = 6" and equal side " (a) = 5" (Given)"$ We know that, $"We know that,"$ Perimeter=(2⋅equal Side)+third Side $"Perimeter" = (2 * "equal Side") + "third Side"$ =(2⋅a)+b $= (2 * a) + b$ =(2⋅5)+6 $= (2 * 5) + 6$ =16 $= 16$ We know that, $"We know that,"$ Area=12⋅base⋅height $"Area" = 1/2 * "base" * "height"$ =12⋅b⋅√a2-b24 $= 1/2 * b * sqrt(a^2 - b^2/4)$ =12⋅6⋅√52-624 $= 1/2 * 6 * sqrt(5^2 - 6^2/4)$ =12⋅6⋅4 $= 1/2 * 6 * 4$ =12 $= 12$

Problem : 13 / 13 [ SectorSegment ] Enter your problem 13. I know that for a sector & segment Radius = 14 and angle of measure = 90 . From this find out length of arc of the sector & segment.


Solution:	Here r=14 and θ=90 (Given) $"Here "r = 14" and " theta = 90" (Given)"$ Length of the arc =l=πrθ180 $"Length of the arc " = l = (pi r theta)/180$ =3.1416⋅14⋅90180 $= (3.1416 * 14 * 90)/180$ =21.98 $= 21.98$ Area of a minor sector =πr2θ360 $"Area of a minor sector "= (pi r^2 theta)/360$ =3.1416⋅142⋅90360 $= (3.1416 * 14^2 * 90)/360$ =153.86 $= 153.86$