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2 years ago

476 Marks boole Calculate the area of this triangle. 17 cm 8 cm 15 cm

Mathematics

1 answer:

Naddika [18.5K]2 years ago

4 0

Answer:

Step-by-step explanation:

sides =8 cm, 17 cm and 15 cm

Area of triangle by Hero's Formula,

S=a+b+c/2

S=17+8+15/2

S=40/2

S=20

A= $\sqrt{s(s-a)(s-b)(s-c)}$

A= $\sqrt{20(20-17)(20-8)(20-15)}$

A= $\sqrt{(20)(3)(12)(5)}$

A= $\sqrt{3600}$

A=60 $cm^{2}$

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Determine if the triangles in the figure are congruent

valkas [14]

Answer:

The triangles are congruent by Angle Side Angle congruency statement and the reason is below.

Step-by-step explanation:

Given:

∠ VUW ≅ ∠ XYW

VW ≅ YW

To Prove:

Δ VUW ≅ Δ XYW

Proof:

In Δ VUW and Δ XYW

∠ VUW ≅ ∠ XYW ……….{Given}

VW ≅ YW ............{Given}

∠ VWU ≅ ∠ XWY …………..{Vertically Opposite Angles are equal}

Δ VUW ≅ Δ XYW .….{ By Angle-Side-Angle congruence test}

......Proved

6 0

2 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

Five minutes on the clock

Nuetrik [128]

Answer:

15/32 cm²

Step-by-step explanation:

Area of rectangle = l × b

=> 3/4 × 5/8

=> 15/32

Therefore,

Area of rectangle is 15/32 cm²